Cracking the Mental Math Code:
A Repertoire of Number Sense Techniques
By: Yash Chandak
Special thanks to
Mr. Newton,
Mom,
Dad,
and Sneha
About this Book
From those who read a poster asking them to join the school’s Number Sense team to the seasoned competitor eyeing State medals, this book serves as a guide and as inspiration, to not only pique interest but to build a foundation on.
You’ll find in this book a stepbystep guide on howto solve Number Sense problems and prepare you for any mental math competitions. There are over 80 mental math techniques are explained in this book, and over 750 problems to learn, train, and improve with. Just a note: as fun as it is to learn new tricks and techniques, it’s just as important to master them and implement them in fulllength tests!
I wish you all the best of luck, and happy computing!
Contents
1 [+ PEMDAS+]
1.1 Fast Addition and Subtraction
1.2 PEMDAS Approximations
1.3 Multiply by 11 and like
1.4 Multiplying by 25
1.5 Multiplying by 125
1.6 Multiplying by any Number Ending in 5
1.7 Multiplying by 101 and like
1.8 Multiplying Two Numbers just above 100
1.9 Multiplying Two Numbers just below 100
1.10 Multiplying 2 Numbers with an Even Difference
1.11 Multiplying 2 Numbers with the Same Tens Digit and Ones Digits that Add to 10
1.12 FOIL
2 [+ Squares+]
2.1 Squares
2.2 Squares Ending in 5
2.3 Sum of Squares in the form a^{2}[++(3a)+]^{2}
2.4 Sum of Two Consecutive Squares
2.5 Sum of Squares when the Outer Digits add to 10 and the Inner Digits are 1 Apart
2.6 Approximations of Exponents
3 +] [+GCD and LCM
3.1 How to find GCD
3.2 How to find LCM
4 [+ Remainder+]
4.1 Remainder with Division
4.2 Remainder with Operations
4.3 Remainder with Exponents
5 [+ Higher Order Exponents+]
5.1 Cubes
5.2 Powers of Special Numbers
5.3 Difference of Cubes 1 Apart
5.4 Laws of Exponents
6 [+ Roots+]
6.1 Square Roots
6.2 Nested Square Roots
6.3 Other Roots
6.4 Approximations of Roots
7 [+ Primes and Divisors+]
7.1 Primes
7.2 Number of Positive Integral Divisors
7.3 Sum of Positive Integral Divisors
7.4 Relatively Prime
8 [+ Other Topics+]
8.1 Additive and Multiplicative Identities and Inverses
8.2 Absolute Value
8.3 Roman Numerals
8.4 Conversion Factors
8.5 Complex Numbers
9 [+ Types of Numbers+]
9.1 Polygonal Numbers
9.2 Deficient, Perfect, and Abundant Numbers
9.3 Happy and Unhappy Numbers
9.4 Evil and Odious Numbers
9.5 Polite Numbers and Politeness
9.6 Frugal, Economical, Equidigital, Wasteful, and Extravagant Numbers
10 Factorials and Combinations
10.1 Factorials
10.2 Permutations
10.3 Combinations
11 Sequences
11.1 Consecutive Integer Sequences
11.2 Arithmetic Sequences
11.3 Geometric Sequences
11.4 Fibonacci
11.5 Other Sequences
12 Memorization
12.1 Fractions and Decimals
12.2 Approximations
13 +] [+Adding Fractions
13.1 Adding Inverses
13.2 Fractions in the form a/b+b/(a+b)
13.3 Special Fractions
13.4 Adding Fractions with no Trick
14 +] [+Multiplying Fractions
14.1 Whole Numbers Same and Fractions Add to 1
14.2 Multiplying 2 Mixed Numbers with the Same Fraction
14.3 Fractions in the Form a*a/(a+b)
14.4 Fractions in the Form a*(a+n)/(a+2n)
14.5 Multiplying using Improper Fractions
14.6 Comparison of Fractions
15 +] [+Repeating Decimals
15.1 Repeating Decimals in the Form of .aaa…, .ababab…, or .abcabcabc…
15.2 Repeating Decimals in the Form of .abbb…, .abccc…, or .abcbcbc…
16 +] [+Bases
16.1 Base n to Base 10
16.2 Base 10 to Base n
16.3 Base a to Base a^n
16.4 Decimals in Other Bases
17 +] [+Linear Equations
17.1 Solving Linear Equations
17.2 Slope and yintercept
17.3 Inequalities
18 +] [+Quadratic Equations
18.1 Roots
18.2 Vieta’s Formulas
19 +] [+Sets
19.1 Subsets
19.2 Proper and Improper Subsets
19.3 Intersection and Union of Sets
19.4 Cartesian Product
20 +] [+Logarithms
20.1 Solving Equations with Logarithms
20.2 Nested Logarithms
21 +] [+Matrices
21.1 Adding and Subtracting Matrices
21.2 Multiplying Matrices
21.3 Determinants
Answers
[*1.1 *]Fast Addition and Subtraction
Most all tricks in this book will require you to be quick in basic operations. Given that you only have 7.5 seconds to read, solve, and write the answer for each question, you don’t get much time for the intermediate steps. So, let’s teach you a trick on arguably the simplest topic in mathematics: addition and subtraction.
Basic addition involves adding one digit at a time moving right to left. To get the answer faster, however, you can add two digits at a time, write the answer for that, and maintain the carries. Let’s see a couple of examples.
Note: it’s almost always advantageous to subtract before you add. Dealing with smaller numbers is usually easier.
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[*1.2 *]PEMDAS Approximations
Almost every Number Sense test will have an approximation question that is solely addition and subtraction, usually 5 or 6 digits long. Since you can be within 5% of the actual answer and still get the question correct, it is advantageous for you to only do the operations for the highest 2 place values and round appropriately rather than trying to chug through all the digits.
To solve this question, I would use the tenthousands and thousands digits to determine an approximate value.
My answer: 96,000
Range: 91,545 to 101,181
Most times, rounding in these questions is beneficial because the ranges are quite large and the problem can become quite easier to solve. However, if you round up on one number a substantial amount (relative to the number of course), it’s good to round down on another to balance out the error.
It’s important to be both quick and accurate when rounding. You don’t want to waste time on easy questions trying to choose how to round. Just do what feels right and start calculating!
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[*1.3 *]Multiply by 11 and like
The one thing more common than addition on a Number Sense test is multiplication. I cannot emphasize how drastically multiplication speed affects your score. It goes without saying that with a lot of multiplication comes a lot of multiplication tricks. The easiest and most common trick is multiplying by 11.
I could try to explain this trick in words for you, but that would be quite confusing and lengthy, so let’s use an example.
Let’s use the original way of multiplying to see the trick.
^. p. 
^. p. 3 
^. p. 5 
^. p. x 
^. p. 1 
^. p. 1 
^. p. 
^. p. 3 
^. p. 5 
^. p. 3 
^. p. 5 
^. p. 0 
^. p. 3 
^. p. 8 
^. p. 5 
It looks like all we did was add the two digits (3+5=8) and put that sum in the middle of that number. Let’s see another example.
^. p. 
^. p. 5 
^. p. 7 
^. p. x 
^. p. 1 
^. p. 1 
^. p. [] 
^. p. 5 
^. p. 7 
^. p. 5 
^. p. 7 
^. p. 0 
^. p. 6 
^. p. 2 
^. p. 7 
1
1
In this example, the sum of the two digits was 12, which sadly isn’t just one digit. So, the 2 was put between the two numbers again but the 5 was increased to a 6 because of the carry in 12.
Using the previous two examples, we can create the following steps to follow when multiplying by 11:
Step 1: write the ones digit as the ones digit of your answer.
Step 2: add the ones digit and the tens digit. If the value is less than ten, write the digit as the tens digit of your answer. Else, note the carry and then write one digit.
Step 3: write the tens digit (after adding any carries) as the hundreds digit of your answer.
Let’s see it pictorially. Each line in this diagram represents the addition for the trick.
^. p=. 5 
^. p=. 7 
^. p. 
^. p. 
^. p. 
^. p<>. Ones 
^. p. 
^. p. 
^. p<>. Tens 
^. p. 
^. p. 
^. p<>. Hundreds 
This trick even works for numbers with more than 2 digits. We keep adding 2 digits at a time, working from right to left.
Step 1: write the 9.
[*Step 2: *]add 2 and 9 to get 11. Write 1 and carry 1.
[*Step 3: *]add 4 and 2 to get 6. Add the carry to get 7. Write 7.
[*Step 4: *]write 4. Our final answer is 4719. Checking it with basic multiplication, we see not only the answer but the same addition we did to get the answer.
^. p. 
^. p. 4 
^. p. 2 
^. p. 9 
^. p. 
^. p. 
^. p. 1 
^. p. 1 
^. p. 
^. p. 4 
^. p. 2 
^. p. 9 
^. p. 4 
^. p. 2 
^. p. 9 
^. p. 0 
^. p. 4 
^. p. 7 
^. p. 1 
^. p. 9 
This trick can also be extended to multiply by 111. When multiplying by 111, you follow the same steps as multiplying by 11 but build up to adding 3 digits at a time.
Step 1: write the last digit, a 9.
Step 2: add 2 and 9 to get 11. Write the 1 and carry 1.
Step 3: this is where it changes. Instead of shifting over to add just 4 and 2, we’re gonna add the 4, 2, and 9 together along with the carry to get 16. We did this because there are 3 ones in the number we are multiplying with. Rule of thumb: the number of ones is the same as the number of digits you have to add.
Step 4: add the 4 and 2 as well as the carry to get 7.
Step 5: write the 4. Final answer: 47619
^. p=. 2 
^. p=. 9 
^. p. 
^. p. 
^. p. 
^. p<>. Ones 
^. p. 
^. p. 
^. p<>. Tens 
^. p. 
^. p. 
^. p<>. Hundreds 
^. p. 
^. p. 
^. p<>. Thousands 
^. p. 
^. p. 
^. p<>. Ten thousands 
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[*1.4 *]Multiplying by 25
Multiplying by 25 is not only quite common but also quite easy. The easiest way to visualize this trick is with money. It’s known that 4 quarters make a dollar and each quarter is worth 25 cents, so let’s use that. If we multiply 25 by 4, we get 100, an easy number to work with. However, we have to divide the other number by 4 to keep the transformed expression equal. Let’s look at an example.
Sometimes, the other number is not divisible by 4. When the number isn’t divisible, imagine the decimal remainder that would occur. When this is subsequently multiplied by 100, shift the decimal place over.
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[*1.5 *]Multiplying by 125
Multiplying by 125 is quite similar to Multiplying by 25 (Trick 1.4). Since, we can divide the other number by 8 and then multiply by 1000.
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[*1.6 *]Multiplying by any Number Ending in 5
The number 5 is so versatile. One of the helpful things when conducting mental math is that. With this obvious fact in mind, let’s look at the doubleandhalf trick. The name says it all; double the number ending in 5 and half the other one. This will keep the value of the expression the same but make the problem much easier to solve.
If the other number is odd, it won’t divide evenly by 2. In this case, it is faster to FOIL (Trick 1.13).
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[*1.7 *]Multiplying by 101 and like
It is very difficult to go wrong when multiplying by 101 if you know what you’re doing. The most important thing to notice is that multiplying by 101 is the same thing as multiplying by 100, multiplying by 1, and adding the two products. Let’s see an example with a 2digit number.
The quotient is literally the number twice, which makes sense given the operations we did. This works for all two digit numbers. Now let’s up the game.
With three digits, we had to add the hundreds digit from the product with 1 and the ones digit from the product with 100. So, when solving multiplication with 101, be wary of addition when it’s not with a twodigit number.
The same theory applies for multiplying by 1001. Just be wary of the place values when thinking of the addition. Rather than writing out an explanation, I’ll let you figure out the place values with practice problems.
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[*1.8 *]Multiplying Two Numbers just above 100
Let’s prove the trick of how to multiply two numbers above 100.
Well, that doesn’t look like a trick. But, let’s look at it a bit closer. The 10000 won’t affect any place value below the tenthousands place and the term won’t affect any place value below the hundreds place. That means that the term is the only term that affects the tens and ones place. Similarly, the term will affect the thousands and hundreds place (along with carries from the term). Let’s see an example to see this in action.
Step 1: Multiply the amount above 100 for both numbers and make it take up two digits. Multiply 2 and 3 to get 06.
Step 2: Add the amount above 100 for both numbers and make it take up two digits. In this problem, add 2 and 3 to get 05.
Step 3: Write a 1. Final answer: 10506.
Let’s try a more complicated version of this problem.
Step 1: Multiply 15 and 9 to get 135. Write 35 and carry 1.
Step 2: Add 15 and 9 along with the carry of 1 to get 25.
Step 3: Write a 1. Final answer: 12535.
This trick also works for 2 numbers just above any multiple of 100 as long as the multiple is the same.
When comparing this formula to the one for just above 100, we see that is being multiplied by the term in Step 2 and is the ten thousands place. Let’s do an example.
Step 1: Multiply 11 and 6 to get 66. Write 66.
Step 2: Add 11 and 6 to get 17. Multiply this by 5 to get 85.
Step 3: calculate 5^{2} to get 25. Final answer: 258566.
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[*1.9 *]Multiplying Two Numbers just below 100
This trick is very similar to Trick 1.7, from derivation to operations (for obvious reasons). But, since it’s important to know, it’s in this book.
If you grasped Trick 1.7, the first three steps in this derivation probably make sense. Notice how I didn’t multiply the two integers and rather factored out the 100 from three terms. That allowed me to get to step 4, which seems like a good stopping point. But why go one step further? Look at the original expression. Now, look at what’s in square brackets. See any similarities?
The term in the square brackets is one of the numbers you will see on your test. When trying to derive tricks with algebraic terms, it’s important to look back at the original expression to see any similarities or simplifications.
Now, what does the last line say? The last two digits of your answer are determined by multiplying the each of the differences between the number and 100. The thousands and hundreds digit are one of the numbers minus the difference between the other number and 100. Let’s now try an example.
[*Step 1: *]Calculate and.
Step 2: Multiply 7 and 4 to get 28.
Step 3: Subtract 4 from 93 (or 7 from 96) to get 89. Final answer: 8928.
[*Step 1: *]Calculate and.
Step 2: Multiply 12 and 9 to get 108. Write 08 and carry 1.
Step 3: Subtract 12 from 91 (or 9 from 88) to get 79. Add the carry from Step 2 to get 80. Final answer: 8008.
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[*1.10 *]Multiplying 2 Numbers with an Even Difference
This trick is most often seen when two numbers are multiplied that have a difference of two and the middle number ends in 5. Really, it’s using algebra to simplify the difference of two squares as the multiplication of two numbers. For this trick, it’s important to know your squares (which is covered in section 2).
These problems are given as the right side of the equation and should be converted to a difference of two squares.
To go from step 3 to step 4 will require you to know (and, but if you don’t know that we have bigger problems), but this is much faster than working out the multiplication when you start working with squares. This trick works for any two numbers with an even difference.
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[*1.11 *]Multiplying 2 Numbers with the Same Tens Digit and Ones Digits that Add to 10
I find these problems to be a special case of Trick 1.9. Even though trick 1.9 is fast, this is faster. Let’s see it.
Let a represent the tens digit of both numbers, let represent the ones digit of the first number, and let represent the ones digit of the second number. Then:
Translating the last line to words, we see that the last two digits of the answer is simply the product of the two ones digits and the hundreds digit of the answer is the tens digit multiplied by one more than itself.
Step 1: Multiply the ones digits and make it take up two place values. The product of 3 and 7 is 21, so write 21.
Step 2: Multiply the tens digit by one more than itself. The product of 2 and 2+1=3 is 6, so write 6. Final answer: 621.
Step 1: multiply the ones digits and make it take up two place values. The product of 1 and 9 is 9, so write 09.
Step 2: multiply the tens digit by one more than itself. The product of 12 and 12+1=13 is 156, so write 156. Final answer: 15609.
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[*1.12 *]FOIL
This, hands down, is the greatest and most useful trick in this book. Put in one sentence, this trick makes it possible to multiply any two numbers either 2 or 3 digits in your head. Now you must be asking, “Why mention all the tricks before this if we have FOIL?” Well, every trick before this, based on numerous students as well as personal experience, is faster than FOIL. There are other multiplication tricks you can find online or in other sources, but I found those to be slower or equal to FOIL in terms of time. Let’s see what FOIL says.
Let a represent the tens digit and b the ones digit of the first number and c represent the tens digit and d the ones digit of the second number. That means that the numbers can be represented as the following:
Now, if we want to multiply the two numbers, we can expand it as so:
What does this mean?
[*Step 1: *]The ones digit of the answer is simply the product of the ones digits of the two numbers multiplied.
Step 2: The tens digit of the answer is simply the sum of the product of the Outer digits plus the product of the Inner digits.
Step 3: The hundreds digit is simply the product of the tens digits of the two numbers multiplied.
All the while, follow carries like every other trick before this. Let’s see FOIL in action.
[*Step 1: *]Calculate the ones digit.; write 6 as ones digit and carry 1.
Step 2: Multiply the outer and inner digits to get and. Now add the two products and the carry to get 45. Write 5 as the tens digit and carry 4.
Step 3: Multiply the tens digits.; add the carry to get 14. Final answer: 1456.
[*Step 1: *]Calculate the ones digit.; write 8 as ones digit.
Step 2: Multiply the outer and inner digits to get and. Now add the two products and the carry to get 17. Write 7 as the tens digit and carry 1.
Step 3: Multiply the tens digits.; add the carry to get 3. Final answer: 378.
This also works for 3 digit multiplication; simply group the tens and hundreds digits as one number and FOIL. There are more calculations that happen so it takes longer to solve, but with practice this multiplication becomes easier.
[*Step 1: *]Calculate the ones digit.; write 8 as ones digit and carry 1.
Step 2: Group the hundreds and tens digit of each number. Multiply the outer and inner numbers to get and. Now add the two products and the carry to get 208. Write 8 as the tens digit and carry 20.
Step 3: Multiply the outer numbers (use a multiplication trick or FOIL again).; add the carry to get 425. Final answer: 42588.
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[*2.1 *]Squares
This is a list of squares you should memorize for Number Sense.
^{. p<>. 19[^2}]=361  ^{. p<>. 27[^2}]=729  ^{. p<>. 35[^2}]=1225  ^{. p<>. 43[^2}]=1849 
^{. p<>. 20[^2}]=400  ^{. p<>. 28[^2}]=784  ^{. p<>. 36[^2}]=1296  ^{. p<>. 44[^2}]=1936 
^{. p<>. 21[^2}]=441  ^{. p<>. 29[^2}]=841  ^{. p<>. 37[^2}]=1369  ^{. p<>. 45[^2}]=2025 
^{. p<>. 22[^2}]=484  ^{. p<>. 30[^2}]=900  ^{. p<>. 38[^2}]=1444  ^{. p<>. 46[^2}]=2116 
^{. p<>. 23[^2}]=529  ^{. p<>. 31[^2}]=961  ^{. p<>. 39[^2}]=1521  ^{. p<>. 47[^2}]=2209 
^{. p<>. 24[^2}]=576  ^{. p<>. 32[^2}]=1024  ^{. p<>. 40[^2}]=1600  ^{. p<>. 48[^2}]=2304 
^{. p<>. 25[^2}]=625  ^{. p<>. 33[^2}]=1089  ^{. p<>. 41[^2}]=1681  ^{. p<>. 49[^2}]=2401 
^{. p<>. 26[^2}]=676  ^{. p<>. 34[^2}]=1156  ^{. p<>. 42[^2}]=1764  ^{. p<>. 50[^2}]=2500 
[*2.2 *]Squares Ending in 5
There’s a neat little trick to find the square of any number ending in 5.
The algebra indicates that the last two numbers of the square is a 25. The other digits are simply the tens digit multiplied by one more than itself. For example:
[*Step 1: *]Write 25.
Step 2: Multiply 3 by 3+1=4 to get 12. Final answer: 1225.
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[*2.3 *]Sum of Squares in the form a^{2}+(3a)^2^
The sum of squares in this form is seen on basically every other Number Sense test. It can be solved using simple algebra.
What this means is that you only have to calculate the smaller square and put a 0 after the term. Let’s see an example.
Although significantly less common, questions may arise that are in the form of,, or . Each of these cases has a simple trick that I’ll let you derive.
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[*2.4 *]Sum of Two Consecutive Squares
The sum of consecutive squares can often be computed manually by highscoring individuals. However, this often takes upwards of 10 seconds and breaks tempo. This trick simplifies the process and makes sure you don’t have to calculate any square.
Let’s compare the last line to the first. Both numbers in the original expression are present in the final one, multiplied. In words, this trick says, “The sum of consecutive squares is twice the product of the two base numbers plus one.” All you have to do is multiply the two numbers you see on the problem, double it, and add 1. Example time!
More often than not one of the numbers will end in 5, making the multiplication all the more easier (refer to trick 1.5).
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[*2.5 *]Sum of Squares when the Outer Digits add to 10 and the Inner Digits are 1 Apart
This sum of squares trick has a complicated derivation (as compared to other tricks) but comes up frequently on tests. Usually, this problem can be identified because the squares are ridiculous by themselves to compute. It is important to note that order matters in this trick as the inner and outer digits would change if the order of the squares change.
First, let a represent the tens digit and b the ones digit of the first number. That means that the first number can be represented as the following:
Now, if the inner digits differ by 1, then the tens digit of the second number is 1 less than the ones digit of the first number. Similarly, if the outer digits add to 10, then the ones digit of the second number is 10 less than the tens digit of the first number. This can be written as the following:
Now, we’ll use algebra to square both expressions and add the terms.
As can be seen by the final expression, we will need to multiply by 101 (refer to 1.6 if you need a refresher) and calculate squares of individual digits, more specifically the sum of the squares of the digits of the first number. As said before, order matters; the first number must have a ones digit one greater than the tens digit of the second number. Let’s see an example.
[*Step 1: *]make sure that the outer digits add to 10 (5+5) and the inner digits have a difference of 1 (76)
Step 2: calculate the sum of the squares of the digits of the first number ()
[*Step 3: *]multiply this sum by 101 to get the final answer (7474)
It is crucial to note that this trick will not work for because. Almost every time the question asks the sum of two large squares, the numbers follow this pattern.
On a rare occasion, you may need to reorder the two squares to see the pattern. For example, if this question read, you would need to switch the position of the two squares before continuing with Step 1.
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[*2.6 *]Approximations of Exponents
Approximating squares and cubes is essential for Number Sense. One thing to look for is a number you know the square to that you can obtain by light rounding. Beware that excessive rounding with base numbers should be avoided because when the number is squared, the minor change becomes major.
Sometimes using properties of exponents help make an approximation question manageable. One that comes into use is:
Hopefully these practice problems help you build speed and see how to manipulate the numbers.
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[*3.1 *]How to find GCD
The concept of Greatest Common Divisor (or Greatest Common Factor) and Least Common Multiple will require knowledge in prime numbers (covered in Chapter 7). Most students in middle school and higher know the conventional way to calculate the two values: compare the prime factorization of the two numbers. Though accurate, this method isn’t efficient because it requires a student to find the complete prime factorization of both numbers, compare them, and multiply the numbers that are common.
Rather, there is simpler way to find GCD when working with 2 numbers. It can be shown that the GCD of two numbers divides their difference. That means that when you find the positive difference of two numbers, the GCD must be a factor of that number.
Let be the GCD of and, which are defined as positive integers (numbers with no fractional parts) and. Also, let and be positive integers. It can then be written that:
Now, let’s subtract both numbers.
We see that the GCD is a factor of the positive difference. That means that instead of using the traditional way of finding GCD, we can subtract the numbers and use this number to find the GCD. After finding the difference, use simple trial and error starting from the largest factor to find the GCD. Although still a bit of work, this is much easier and faster. Usually the difference doesn’t have many factors and the GCD becomes obvious. Let’s see some examples.
I used the prime factorization of 51 to realize that both numbers are a multiple of 17.
Once again, I used the prime factorization of 46 to analyze the original two numbers and determine the GCD.
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[*3.2 *]How to find LCM
Another interesting property of GCD and LCM is the following:
I’ll explain this formula with a Venn diagram.
As seen in the diagram above, the two circles represent the two numbers. The intersection would be the GCD as it divides into both numbers. What’s left is each number divided by the GCD. Now, to find the LCM, we need to get the smallest number that divides both numbers. Pictorially, we need to cover each part of the diagram once.
To find LCM, we can find GCD using trick 3.1, find the product of the two numbers, and then divide the latter by the former (although it is advisable to divide before to multiply, so divide one of the numbers by the GCD and then multiply by the other number). Let’s use an example from 3.1.
It was determined that:
Additionally, questions sometimes arise that give you 3 of the 4 numbers in the last equation and ask for the fourth one. Knowing this trick, such questions become a breeze.
To find x, simply use the expression:
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[*4.1 *]Remainder with Division
Number Sense usually asks you to find the remainder when dividing by 2 numbers: 9 and 11. The proofs of each trick involves modular arithmetic and isn’t necessary to know. Instead, let’s focus on the trick itself.
The remainder when a number is divided by 9 is the same as the remainder when the sum of the digits of the number is divided by 9. That means you can add the digits of the number and divide that by 9 rather than the original number. Let’s see an example.
However, there’s a way to make the thought process easier. Every time you get 9 in your addition, make it 0. In the previous example, you would likely start left to right. , and Now that you have a multiple of 9, make it 0. Moving forward, so your final remainder is 8.
To find the remainder when dividing by 11, we need to employ a slightly more complex trick.
Step 1: start at the ones digit and add every other digit reading backwards (right to left). This will be called.
Step 2: start at the tens digit and add every other digit reading backwards (right to left). This will be called.
Step 3: Determine. If this difference is between 0 and 10, write the answer. If the difference is less than 0, add multiples of 11 until you get into this range. If the difference is greater than 10, subtract multiples of 11 until you get into this range.
The reason you need to get into the range of 0 to 10 is because when dividing an integer by 11, the only possible remainders are 0 to 10.
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[*4.2 *]Remainder with Operations
This trick is quite useful to drastically simplify questions asking about remainder. When trying to find the remainder of an expression divided by a divisor, instead of using the original terms in the expression, you can use the remainder when each individual term is divided by the divisor. What does that mean?
Step 1: Find the remainder of each term when divided by the divisor.
Step 2: Plug in each remainder for the original number and keep the operations the same.
Step 3: Calculate the final remainder.
Note that the quotient of the original expression and the final expression are not the same, only the remainder (which is what we’re looking for).
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[*4.3 *]Remainder with Exponents
Remainders with exponents are quite similar to Trick 4.2. Usually a late fourth column question, these questions can be extremely simple or fairly difficult. The entire trick is attempting to establish a pattern. There are 2 common, easy patterns that I’ll mention, but the others will require you to do multiplication. As a side note, you should be familiar with the Laws of Exponents (Trick 5.4).
When determining the remainder of an expression in the form of:
If, first find the remainder of (let this be). The remainder of will have the same remainder as. This means that you can plug in the remainder rather than the original number and still get the same remainder (just like in Trick 4.2).
Pattern 1: When the base of the exponent is one greater than the dividend.
In this pattern, the answer will always be 1.
Pattern 2: When the base of the exponent is one less than the dividend.
This will take the form of:
We know for a fact that divided by will have a remainder of since it will not divide into a number one greater than itself. We also know that:
We can see that is divisible by, so will have a remainder of 1 when divided by.
Looking at the pattern, if the exponent is even, the remainder will be 1. If the exponent is odd, the remainder will be the base number (in the example it’s 17).
[*Else: *]If neither of the two conditions are met, keep plugging in the remainder to try to either get a pattern or get to the answer. For example:
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[*5.1 *]Cubes
Here are a list of cubes you need to memorize.
^{. p. 4[^3}]=64  ^{. p. 7[^3}]=343  ^{. p. 10[^3}]=1000  ^{. p. 13[^3}]=2197 
^{. p. 5[^3}]=125  ^{. p. 8[^3}]=512  ^{. p. 11[^3}]=1331  ^{. p. 14[^3}]=2744 
^{. p. 6[^3}]=216  ^{. p. 9[^3}]=729  ^{. p. 12[^3}]=1728  ^{. p. 15[^3}]=3375 
[*5.2 *]Powers of Special Numbers
You also want to memorize these numbers. Note that oftentimes these powers will be combined in approximation questions.
^{. p<>. 3[^1}]=3  ^{. p<>. 5[^1}]=5  ^. p<>. π^1^≈3.1 
^{. p<>. e[^1}]≈2.7  ^. p<>. ϕ^1^≈1.6 
^{. p<>. 3[^2}]=9  ^{. p<>. 5[^2}]=25  ^{. p<>. π} 2^≈9.9  ^{. p<>. e[^2}]≈7.5  ^{. p<>. ϕ} 2^≈2.6 
^{. p<>. 3[^3}]=27  ^{. p<>. 5[^3}]=125  ^{. p<>. π} 3^≈31  ^{. p<>. e[^3}]≈20  ^{. p<>. ϕ} 3^≈4.2 
^{. p<>. 3[^4}]=81  ^{. p<>. 5[^4}]=625  ^{. p<>. π} 4^≈97  ^{. p<>. e[^4}]≈55  ^{. p<>. ϕ} 4^≈6.8 
^{. p<>. 3[^5}]=243  ^{. p<>. 5[^5}]=3125  ^{. p<>. π} 5^≈306  ^{. p<>. e[^5}]≈150  ^{. p<>. ϕ} 5^≈11 
^{. p<>. 3[^6}]=729  ^{. p<>. 5[^6}]=15625  ^{. p<>. π} 6^≈960  ^{. p<>. e[^6}]≈400  ^{. p<>. ϕ} 6^≈18 
^. p. 
^. p. 
^{. p<>. π} e^≈22.5  ^. p<>. e^π ^≈23 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
Oftentimes, approximating π^2^ to be 10 and π^4^ to be 100 works if you subtract a little bit at the end. It’s also convenient to know e^{3} is 20 and work from there.
[*5.3 *]Difference of Cubes 1 Apart
Usually the difference of two cubes comes late in the fourth column. As with numerous other tricks, finding the difference of cubes 1 apart will require simple algebra. It’s important to know that if you encounter new variations of questions that you want to derive a trick for, it is best to try to start with algebra and work to the original numbers if possible. Let’s take a look at this derivation.
Looking at the final expression, we can conclude that the positive difference of cubes one apart is one more than the product of the two bases tripled. For example:
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[*5.4 *]Laws of Exponents
It’s crucial to know how to work with exponents to simplify a difficult expression. First know that. When the base is the same, the following statements are true:
However, not every question will have numbers with the same base. When this is the case, know the following statements:
Most every time you must manipulate the numbers given to fit one of the expressions above. For this, use the following fact:
Let’s work an example now.
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[*6.1 *]Square Roots
Square roots are the opposite of squares (much like subtraction is the opposite of addition and division that of multiplication). Number Sense will require you to memorize the following square roots to the thousandths place.
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
There are two types of questions asked: truncate and round. When you truncate, you always round down, no matter if the next place value is 1 or 9. In contrast, when you round, you look at whether the next place value is less than 0.5 or not.
[*6.2 *]Nested Square Roots
A square root is a number raised to the ½ power. Nested square roots are intimidating and seemingly complex, but they are simple if you know how to work with the radicals. If given as an expression (as compared to an equation) and asked to find what it equals, work insideout. For example:
However, if given as an equation, you may need to square both sides to denest the nested square root. Let’s find X in the following examples.
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[*6.3 *]Other Roots
Numbers can be raised to other fractional exponents as well. The 1/3 power is called the cube root. These fractional exponents are often simplified using Laws of Exponents (Trick 5.4).
Note how in the second example there are two paths, both of which get to the correct answer. If you can recognize that 1024 is an even power of 2 and thus is a power of 4, the calculations become a bit simpler and quicker.
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[*6.4 *]Approximations of Roots
The majority of questions with roots on a Number Sense test deal with approximations. There are a handful of strategies to quickly and accurately estimate roots.
Step 1: Look for a perfect square that is close to the number under the radical.
[*Step 2: *]Employ the following fact:
For example:
Step 3: For long square roots, only worry about the most important digits; that is, ignore the digits in the tens and ones place. This step is best explained with an example.
Note how we can round the tens and ones digits such that we can factor out a 100. Now, we just worry about the 145.
Note that the actual answer is 120.58 and the range of accepted answers is 115127. Let’s try another example.
Since the radical is still fairly large, we can factor out one more 10 and not affect the value too much.
The correct answer is 1110.54 with a range of 10551166. In this example, since we rounded 1233321 down to 1210000, it wouldn’t do much harm to add 5 or 10 to our final answer. Either way, we would fall in the range.
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[*7.1 *]Primes
A prime number is one that is divisible only by 1 and itself. Every integer greater than 1 can be written as the product of 1 or more primes; this is called a number’s prime factorization. Since prime numbers are essential in competition math (including Number Sense), here is a list of primes under 100 you should be familiar with.
^. p=. 13 
^. p=. 31 
^. p=. 53 
^. p=. 73 
^. p=. 17 
^. p=. 37 
^. p=. 59 
^. p=. 79 
^. p=. 19 
^. p=. 41 
^. p=. 61 
^. p=. 83 
^. p=. 23 
^. p=. 43 
^. p=. 67 
^. p=. 89 
^. p=. 29 
^. p=. 47 
^. p=. 71 
^. p=. 97 
Note that neither 0 nor 1 is prime. Also, 2 is the only even prime number as every even integer greater than 2 will be divisible by 2.
[*7.2 *]Number of Positive Integral Divisors
We use the prime factorization of a number to determine the number of positive integral divisors (factors) it has. It’s best shown with an example.
Using the prime factorization, we can see that any combination of from to and from to will create a factor of 12.
^. p. 
^. p. 
^. p. 
In fact, we can derive a formula based on the prime factorization of a number that tells us the number of factors the number has. Since had 2 possible combinations ( and) and had 3 possible combinations, we can see that there is one more combination than the exponent of the prime number. If we multiply these combinations, we get the number of factors.
If we define to be where are prime, then:
When trying to determine the prime factorization, make sure to always be working with primes (hint: 4 isn’t prime, contrary to the belief of many students in the middle of a Number Sense test). Keep dividing the number by the largest prime you see it is divisible by. Let’s work an example from start to finish.
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[*7.3 *]Sum of Positive Integral Divisors
The sum of a number’s factors has a fairly lengthy formula but is faster than figuring out each factor and adding them up. We will stick with our definition in Trick 7.2 of to be where are prime. The formula reads:
The formula is quite daunting, but is actually repetitive and goes fast when you have memorized the powers listed in Trick 5.2: Powers of Special Numbers. Let’s work a couple of examples.
If the exponent in the prime factorization is 1, the term will equal. Also, when the prime number is 2, the denominator will always be 1 and can thus be ignored. This will simplify calculations tremendously.
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[*7.4 *]Relatively Prime
Two numbers are relatively prime when their GCD is 1 (or the LCM is the product the two numbers). Almost every question pertaining to relatively prime numbers in Number Sense will read “How many positive integers less than or equal to X are relatively prime to X?”
There’s actually a pretty simple formula for this. Sticking with our definition in Trick 7.2 of to be where are prime, the formula reads:
Notice that this formula doesn’t use the exponents in the prime factorization. The fractions should cancel out and yield an integer, else you probably did something wrong. Let’s see an example:
There are a variety of ways to simplify the multiplication; I’ll leave it up to you to decide what works best for you. Just a word of advice: it’s better to divide or simplify before multiplying.
[*8.1 *]Additive and Multiplicative Identities and Inverses
The additive identity is the number than when added to any value yields the original value. Thus, the additive identify is 0. Similarly, the multiplicative identity is the number that when multiplied to any value doesn’t change the value. So, the multiplicative identity is 1. We use identities to calculate inverses.
A value added to its additive inverse yields the additive identity, or 0. That means that the additive inverse is just the original value with the opposite sign.
A value multiplied to its multiplicative inverse yields the multiplicative identity, or 1. That means that the multiplicative inverse is just the reciprocal of the original value.
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[*8.2 *]Absolute Value
The best way to describe absolute value is the distance between the 0 and the number on a number line. Basically, it makes any number positive (and keeps 0 at 0). Number Sense usually asks nested absolute value questions, that is, absolute values inside absolute values.
The best way to solve nested absolute value questions is to work inside out. Since most all operations in these questions are single digit addition and subtraction, the number one thing you need to watch are the signs.
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[*8.3 *]Roman Numerals
It’s important to be able to read Roman Numerals quickly and accurately. First, you need to know the following conversions:
^. p=. C = 100 
^. p=. D = 500 
^. p=. M = 1000 
^. p. 
To represent numbers with Roman Numerals, work from the biggest to smallest value (M to I). Roman Numerals are additive, so you try to find the largest value is less than the number.
Also, you generally write the biggest value first and descend accordingly.
Next, you need to know that there can never be 4 of any letter in a row. Instead, the trick is to subtract to get that value. To write this in a way that indicates subtraction, you write the smaller number before the larger number (this is the exception mentioned before).
The best strategy when converting Arabic numerals (normal digits) to Roman numerals is to work one place value at a time.
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[*8.4 *]Conversion Factors
You should know both the metric and customary conversion factors for length, area, volume, weight, and temperature.
Distance:
^. p<>. 36 inches = 3 feet = 1 yard 
^. p<>. 5280 feet = 1760 yards =320 rods = 1 mile 
Area:
^. p<>. 640 acres = 1 square mile 
Volume:
^. p<>. 2 pints = 1 quart 
^. p<>. 4 quarts = 1 gallon 
^. p<>. 2 gallons = 1 peck 
^. p<>. 4 pecks = 1 bushel 
Weight:
^. p<>. 2000 pounds = 1 ton 
Temperature:
^. p. 
Metric System Prefixes:
^{. p<>. 10[^9}]  ^. p<>. Nano: 
^{. p<>. 10^9} 
^{. p<>. 10[^6}]  ^. p<>. Micro: 
^{. p<>. 10^6} 
^{. p<>. 10[^3}]  ^. p<>. Milli: 
^{. p<>. 10^3} 
^{. p<>. 10[^2}]  ^. p<>. Centi: 
^{. p<>. 10^2} 
^{. p<>. 10[^1}]  ^. p<>. Deci: 
^{. p<>. 10^1} 
You use the prefixes on base units for every measurement. These include:
It’s also important for you to know the following conversions as they frequently appear on Number Sense tests.
table((((((.
^.
p.

^.
p.

[*8.5 *]Complex Numbers
It is well established that square roots can only be taken of positive numbers. But what about the square root of negative numbers? The square root of a negative number doesn’t result in a real number (since no number squared can be negative), so we define the imaginary number to be. That means that:
The cycle continues for every 4 powers. Problems in Number Sense asking about imaginary numbers will be in the following form:
To solve such questions, you will have to expand the expression by multiplying the two binomials.
Notice how the problem above only asked for, or the real number component of the resulting binomial. Thus, we didn’t have to calculate. When multiplying two binomials in this form, know the following:
, or the real number component of the answer, is determined by multiplying the two real numbers in each binomial and adding this to the product of the two imaginary numbers in each binomial.
, or the coefficient of the imaginary component of the answer, is determined by multiplying the real number of one binomial by the imaginary component of the other binomial and adding the two products.
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[*9.1 *]Polygonal Numbers
Polygonal numbers probably the most common type of number on the test. These can be triangular numbers, squares, pentagonal numbers, and so on. For example, 10 is a triangular number because you can construct a triangle with 10 dots like bowling pins.
However, 10 isn’t a square number because 10 dots cannot be arranged to make a square. Instead, 9 dots can make a square. Each type of polygonal number has its own sequence and are completely separate from one another. However, the formula for calculating the n^{th} sgonal number is the same:
It’s also helpful to be acquainted with triangular numbers as they show up frequently: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, and so on. The formula for triangular numbers is also useful because it is much simpler than the formula above:
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[*9.2 *]Deficient, Perfect, and Abundant Numbers
To determine whether a number is deficient, perfect, or abundant, you look at the sum of the divisors of the number. Every positive integer is either deficient, where the sum of the factors is less than twice the number; perfect, where the sum of the factors is equal to twice the number; or abundant, where the sum of the factors is greater than twice the number.
Just some notes. The only perfect numbers less than 10,000 are 6, 28, 496, and 8128 (which you should memorize). The smallest abundant number is 12, so every single digit integer is either deficient or perfect. Most abundant numbers are even; the first odd abundant number is 945.
[*9.3 *]Happy and Unhappy Numbers
Happy numbers take a fair amount of computation to get to. To determine if a number is happy or not, begin by squaring each digit and adding up the squares. With the number that results, repeat the steps; square each digit and add up the squares. Continue this until one of two things happen:
Therefore the number 23 is happy.
Note how 89 shows up again (step 3). That means that we’re going to cycle with the same numbers infinitely if we continue the calculations. That means that 25 is unhappy.
Now, it is obviously very tedious to calculate all of those squares and do the addition. That’s why it’s best to memorize some of the unhappy numbers so that as soon as you see them, you can stop and conclude that the number is unhappy.
[*9.4 *]Evil and Odious Numbers
To determine if an integer is evil or odious, the first step is to convert the number to base 2, or binary (bases are covered in Chapter 16). After the number is converted to binary, you count the number of ones that appear in the binary form. If the number of ones is even, the number is evil; if odd, then odious (even: evil and odd: odious).
[*9.5 *]Polite Numbers and Politeness
A polite number is any number that can be written as the sum of two or more consecutive integers. The most important takeaway is that powers of 2 are impolite; every other positive integer is polite.
The number of ways a number can be expressed as the sum of positive integers is called the number’s politeness. The politeness of a number is the number of odd divisors the number has. To determine this, follow the following steps:
Step 1: Find the prime factorization of the number
Step 2: Add 1 to every power of a prime number greater than 2
Step 3: Multiply each sum
Step 4: Subtract 1 from the final product
For example:
[*9.6 *]Frugal, Economical, Equidigital, Wasteful, and Extravagant Numbers
To determine if a number is frugal (or economical, as they are the same thing), equidigital, or wasteful (or extravagant, as they are the same thing), first find its prime factorization. Then count the number of digits in the number’s prime factorization (including exponents besides 1) and compare it to the number of digits in the number itself. If:
Take a look at the following examples:
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[*10.1 *]Factorials
The factorial of a positive integer is the product of every integer from the integer down to 1. Factorials are used when calculating how many possible ways there are to arrange objects, probability, and algebra. It is also worth noting that 0! = 1.
For example:
[*10.2 *]Permutations
The idea behind permutations is that you are ordering all the elements of a set, either with replacement or without, while caring about the order they’re put in.
For example, to calculate the number of possible codes of a 3digit lock, you would look at permutations because order matters. If a digit could be used more than once, the permutations would be with replacement (once you use a number, it can be used again).
Since there are 10 possible digits to choose from for each slot, there are:
However, if every digit had to be separate, then there would be no replacement (once you use a number, you can’t use it again).
There are 10 possible digits to choose from for the first slot. For the second slot, since we can’t choose the number chosen in the first slot, there are only 9 possible digits. Likewise, the third slot only has 8 possible digits. This results in:
Let’s say we want to arrange all 10 digits to create a 10 digit number. Conventionally, numbers don’t start with 0, but we’ll accept it for this problem. We have 10 options for the first digit of the number, 9 options for the second, and so on and so forth until there’s 1 option for the last digit. The number of possible 10 digit numbers with these conditions is:
As for notation, to count the number of orderings for a set of elements r that are chosen from the set n, we say it to be:
For instance, the example of creating a 3digit number would be written. When given in this form, you calculate it with replacement. The general formula is:
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[*10.3 *]Combinations
Combinations counts how to choose items from a set in such a way that order doesn’t matter. Many times this is phrased as the combination of n things taken k at a time. Like permutations, combinations can be calculated with or without repetition.
Assume we have a bag of 6 basketballs labeled 1 to 6. If we were to choose 2 basketballs from the 6 without replacement, we wouldn’t put back the first basketball we chose. That means we have 6 options for the first basketball and 5 for the second one.
However, the problem doesn’t end there. Choosing basketballs 3 and 5 is the same thing as choosing basketballs 5 and 3, yet they are counted individually so far. Thus, we have to divide our initial product by 2 to get the possible combinations.
We can continue this process for choosing 3 basketballs from the 6. First, we have 6 basketballs to choose from, then 5, then 4. So the initial product is:
Next, we have to remove repeats. Since choosing basketballs 1, 2, and 3 can also be counted as 132, 213, 231, 312, and 321, we are counting 6 times as many combinations as there really are. To remove these, we divide the product by 6:
We can continue this process for other possible cases. The general formula for choosing r items from a set of n items ends up being:
Note: factorials are easy to cancel out. Always divide before multiplying when working with factorials. For example:
Practice problems including mix between combinations and permutations.
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[*11.1 *]Consecutive Integer Sequences
There are 3 common consecutive integer sequences you should know. The formulas for all three are relatively simple and shouldn’t take much time to calculate.
Sum of consecutive integers starting with 1:
Interestingly enough, this formula is identical to triangular numbers. If you research more into triangular numbers, you can see why this is the case.
Sum of consecutive even integers starting with 2:
Sum of consecutive odd integers starting with 1:
In words, the equations mean:
The sum of consecutive integers from 1 to is the triangular number.
The sum of consecutive even integers from 2 to is half of multiplied by that half plus one.
The sum of consecutive odd integers from 1 to is the square of half of one more than.
Let’s try an example of each.
If a problem arises that doesn’t start with 1 (or 2 for consecutive even integers), add as if the sequence did. Then simply subtract out the missing terms.
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[*11.2 *]Arithmetic Sequences
An arithmetic sequence starts with any number (not integer) and progresses such that the difference between consecutive terms is constant. The most important numbers when working with arithmetic sequences are the first term () and the common difference (). Algebraically, an arithmetic sequence can be expressed as:
Some questions may ask you to find a certain term of the sequence given the first couple of values. Using the initial term and common difference, to find the term plug into the equation:
Let’s see it in action.
Other questions pertaining to arithmetic sequences will ask you add up a certain number of its terms, starting from a value and going until with a common difference. To determine the sum, it should first be noted that:
In words, this says that the sum of the first and last terms is the same as the sum of second and secondtolast terms, which is the same as the sum of the third and thirdtolast terms, etc. Because of this property, the average value of the terms in this finite sequence (number of terms can be even or odd) is:
In this finite sequence there are terms. However in most cases this value isn’t given to us. To find it, use this equation:
Now that we know both the number of terms and the average value of each term, we can find the sum of the terms in the finite arithmetic sequence by multiplying the two quantities:
Since the algebra may look daunting, how about a couple of examples?
Finding the sum of an arithmetic sequence is calculation heavy. It’s always beneficial to divide before you multiply, but the best way to get faster is to work practice problems and understand what works best for you. Let’s work a slightly more complex sequence:
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[*11.3 *]Geometric Sequences
Similar to arithmetic sequences who can have any number as the first term and every consecutive term has a common difference, a geometric sequence can also have any number as the first term and every consecutive term has a common ratio. Instead of adding the common difference to get the next term, you multiply the common ratio. Algebraically, this can be expressed as:
Note that and can be any number (fractional, negative, irrational, etc.). To find the term of a geometric sequence, plug into the following equation:
Geometric sequences go on infinitely. When the ratio is greater than 1 (), the terms will tend to infinity. However, when the ratio is less than 1 (), the terms will tend to 0. In the latter case, there exists a formula for the sum of all terms in this sequence.
Given a sequence that starts with a value and has a common ratio with, the sum of all terms in this sequence is:
Let’s work with a couple of sequences.
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[*11.4 *]Fibonacci
The Fibonacci sequence is a sequence of numbers where the next number is the sum of the previous two terms. The original sequence starts with 0 and 1. The series begins:
It is advisable to be familiar with the first 10 to 15 terms of the sequence. The convention for denoting Fibonacci numbers is and.
Fibonaccilike sequences are sequences that have two different numbers to start but follow the same convention of adding the previous two terms to get the third term. A common example is the Lucas sequence:
Most questions pertaining to Fibonacci or Fibonaccilike sequences are seen in the fourth column of Number Sense. Rather than asking a specific term from a sequence, the question usually asks the sum of a certain number of terms. This can be solved in a variety of ways.
This formula states that the sum of the first terms of a Fibonacci sequence is the sum of twice the value of the Fibonacci term 2 terms ahead and the Fibonacci term 1 term ahead minus the second Fibonacci term. I understand this may be confusing; maybe an example will help.
This formula works best when you know the last two numbers in the sequence (as compared to just the last number in the sequence). Other common questions ask you the sum of the first 8, first 9, first 10, or first 11 terms in a Fibonaccilike sequence. There are simple formulas for each:
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[*11.5 *]Other Sequences
The following sequences seldom show up but are worth knowing.
Squares
Cubes
Triangular Numbers
[*12.1 *]Fractions and Decimals
It is crucial you know the following table so that you can recognize special numbers in problems and convert from one form to another and simplifying the problem.
^. p=. Decimal 
^. p=. Fraction 
^. p=. Decimal 
^. p=. .5 
^. p. 
^. p=. .25 
^. p=. .75 
^. p. 
^. p=. .125 
^. p=. .375 
^. p. 
^. p=. .625 
^. p=. .875 
^. p. 
^. p=. .0625 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p=. .2 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p. 
^. p=. .05 
^. p. 
^. p=. .025 
[*12.2 *]Approximations
Fractions, decimals, and percents are often seen in questions that ask you to approximate the answer; most times they will simplify with other numbers in the expression. The values usually aren’t the exact fraction or decimal but are close enough that you can round and still fall in the range (granted that everything else goes right). The key is to be able to recognize the fraction, convert it to the form you need it to be, and then simplify it with other numbers in the expression.
Here are some examples.
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[*13.1 *]Adding Inverses
Operations with fractions are a big aspect of Number Sense. Luckily, most all questions asked have simple tricks with recurring patterns and simple algebraic proofs, so fear not! Fraction questions are quick and simple, not cumbersome nor lengthy.
The first trick involves adding two fractions, one of which is the multiplicative inverse of the other. First, let’s see the trick:
Note: in the fourth line, we added the value of 0 in the form of. Since adding 0 does not change the value of the numerator, we can use this to help simplify the expression.
Let’s break down the algebraic terms into steps:
[*Step 1: *]write the whole number 2.
Step 2: find the difference between the two numbers and square it (since the difference is squared, it will always be positive). Note this as the numerator.
Step 3: multiply the two numbers together; this becomes the denominator. Simplify with the numerator if possible; write the final fraction next to the whole number.
Time for an example.
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[*13.2 *]Fractions in the form a/b+b/(a+b)
Oftentimes fractions in this form are difficult to recognize and are simple enough to cross multiply (refer to Trick 13.4). However, the trick does save time and thus should be implemented whenever possible.
Let’s break this down.
[*Step 1: *]write the number 1 as the whole number.
Step 2: square the term. Remember this to be the numerator.
Step 3: multiply the denominators of the two fractions to get the denominator of the answer. Simplify with the numerator from Step 2 if possible. Write the final fraction as the fraction next to the 1.
Here’s an example.
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[*13.3 *]Special Fractions
Sometimes in the middle of a Number Sense test you come across a difference of fractions that seems to have no trick. The numbers are large, there’s heavy computation, and time’s dwindling. Luckily, nine times out of ten the question is a special form that makes the calculation significantly easier.
These questions ask about a difference of two fractions where the numerator of the second fraction is a multiple of the numerator of the first fraction plus a certain value. The denominator of the second fractions is also the same multiple of the denominator of the first fraction minus that same certain value used in the numerator.
Hopefully the algebra makes a bit more sense. We’ll use to represent the multiple and to represent that certain value. In that case, we can represent the question as the following:
With this, we can derive a trick using some cross multiplication.
Let’s use the formula in a handful of examples.
Step 1: realize the following:
[*Step 2: *]calculate. In this example.
[*Step 3: *]Add the numerator and denominator of the first fraction, then multiply by. This will be the numerator of your answer. In this example, we see.
[*Step 4: *]Multiply the denominators of the two fractions to get the denominator of your final answer. For this. Final answer: .
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[*13.4 *]Adding Fractions with no Trick
There come times where no trick is available to add or subtract the fractions given. When this situation arises, you have to add fractions the good oldfashioned way: with least common denominators. The numerators and denominators are usually manageable. Practice is what will make these problems fast.
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[*14.1 *]Whole Numbers Same and Fractions Add to 1
This is probably the simplest trick when multiplying fractions. It will ask you to multiply two mixed numbers, the whole numbers of which are the same and the fractions add to 1. To derive the trick, let’s represent the whole number as and the fraction of the first mixed number as. In that case, the fraction of the second fraction will be. Now it’s algebra time!
Although it doesn’t look like much, the last line is a real game changer. It tells us the following things:
[*1) *]The fraction of the answer is the product of the two fractional components of the mixed numbers.
[*2) *]The whole number of the answer is the product of the whole number in the mixed numbers multiplied by one more than itself.
An example is in order now.
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[*14.2 *]Multiplying 2 Mixed Numbers with the Same Fraction
Multiplication of two such numbers can be simple in itself, but it can go even faster if you understand FOIL with fractions. Let and represent the two whole numbers and represent the fraction shared by both proper numbers. In that case, we can see that:
In most cases, the term results in an integer. This means that the only fraction in the answer is. When working with such questions, follow these steps:
Step 1: Add the two whole numbers and multiply the sum by the fraction shared by both terms.
Step 2: If the result of Step 1 is an integer, add it to the product of the two whole numbers and write this as the whole number of the answer. If the result of Step 1 is a mixed number, add the whole number component to the product of the two whole numbers. Wait to write the whole number until after Step 3 because the result of Step 3 may be greater than 1.
Step 3: Square the fraction. If from Step 2 there is a fractional component still left, add this to the square of the fraction.
Now let’s work some examples to get a feel for the different cases.
Now I understand the second and third examples look quite computational and timeconsuming, but it should be noted that most all questions on Number Sense with the same fraction will follow the first example.
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[*14.3 *]Fractions in the Form a*a/(a+b)
When people usually see fractions in this form, they quickly square the numerator and then go about trying to simplify it with the denominator to get the final answer into a mixed number. Although this may be fast on certain questions, the following trick is sure to be faster (with a bit of practice):
A simple analysis of the final algebraic expression results in the following steps to follow:
Step 1: Calculate the value of
[*Step 2: *]Find; this is the whole number of the answer.
Step 3: Square; this becomes the numerator of the fraction. The denominator is simply the denominator of the fractional term.
Let’s see it in action!
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[*14.4 *]Fractions in the Form a*(a+n)/(a+2n)
Here’s another common, easy fraction’s trick that shows up semifrequently. Questions in this form are fairly easy to recognize and evaluate. Let’s take a look at the algebra below.
Skipping down to the last line, we can extract these steps when solving questions in this form:
Step 1: Determine.
Step 2: Square and multiply by 2 to get the numerator of the fraction. Simplify with the denominator if possible and write the fraction as the fractional part of the answer. If the numerator turns out to be greater than the denominator, make it into a mixed number and carry any values.
Step 3: Subtract from and add any carries to get the whole number of the final answer.
Seems simple enough! The number one part that messes people up is forgetting to double the numerator and then simplify. Since you can’t mark over or erase on tests, you should simplify the fraction in your mind before writing the answer. Let’s see a couple of examples.
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[*14.5 *]Multiplying using Improper Fractions
More like a technique rather than a trick, multiplying using improper fractions is easier than multiplying mixed numbers for the following reasons:
[*1) *]Oftentimes numerators and denominators will cancel out.
[*2) *]It is easier to multiply two integers than FOIL two mixed numbers.
Thus, when you see a question which asks the product of two mixed numbers and there seems to be no trick, try converting the numbers to improper fractions. More than likely something will cancel.
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[*14.6 *]Comparison of Fractions
Nearly every other test will ask you to compare two improper fractions. The general technique is to find the least common denominator and then find the numerator for both fractions; the fraction with the larger numerator is the larger fraction.
Instead of following this lengthy trick, we will employ the same principles but simplify calculations. This is done using crossmultiplication.
The basic idea goes that instead of trying to find the least common denominator, we just find a common denominator, namely the product of the two denominators. But we don’t evaluate this. Instead, we just determine the numerators if this were the denominator and compare those numerators.
In other words, we’re going to multiply the denominator of one fraction by the numerator of the other and then the denominator of the other fraction by the numerator of the first (hence, cross multiply).
Let’s see a few examples to make sense of this all.
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[*15.1 *]Repeating Decimals in the Form of .aaa…, .ababab…, or .abcabcabc…
The two most important pieces of information to know before starting repeating decimals are the following:
Note: the bar above the 1 means that the pattern of digits under the bar repeats continuously.
When you have a repeating decimal where every digit is the same, the fraction will be a multiple of. For example:
When you have a repeating decimal in the form of, the fraction will have a denominator of 99 and a numerator of. For example:
Lastly, repeating decimals in the form of will have a denominator of 999 and a numerator of.
The process can also be reversed, where you’re given a fraction and asked to write the first handful of digits of the decimal expansion. This can go like this:
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[*15.2 *]Repeating Decimals in the Form of .abbb…, .abccc…, or .abcbcbc…
Every fraction has a numerator and a denominator; we’ll learn to determine each value separately.
Determining the denominator of these fractions uses the two pieces of information highlighted at the beginning of Trick 15.1. The number of repeating digits is the decimal is the number of 9’s in the denominator, and the number of digits to the right of the decimal point but do not repeat are the number of 0’s in the denominator (before simplification). Let’s look at the three cases:
Now for the numerator. The first step is to imagine that the decimal is written with a vinculum (the overhead bar). Next, take out the decimal point. Finally, subtract the nonrepeating part of the number from the entire number to get the numerator. Let’s see the three cases again:
Now that we know the formulas, let’s work some problems!
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[*16.1 *]Base n to Base 10
A base is the number of digits that can be used to represent numbers. For example, most all computation is in base 10, called decimal, because we use the digits 09 to represent numbers. Other common bases include base 2, or binary, and base 16, or hexadecimal.
We can only use 0’s and 1’s to represent numbers in binary. However, we need more characters than 09 to represent numbers in hexadecimal. Thus, we use the alphabet: A=10, B=11, C=12 … F=15. For example, a number in base 16 could read 3D2 or A7.
The best way to understand numbers in other bases is through place value. In base 10 (what you’re likely used to), there is the ones place value, then tens, then hundreds, etc. Each subsequent place value is 10 times the previous, where the first place value is 1.
The same pattern occurs in numbers in other bases, where each time you increase a place value you multiply by the base you’re in, starting from 1. Basically, you replace the 10 in the expression above with the base you’re working in. This is how you convert numbers in other bases to base 10.
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[*16.2 *]Base 10 to Base n
Converting a number from base 10 to another base has a completely different system. Here are the steps to follow:
Step 1: Divide the number you want to convert by the desired base you want to convert to.
Step 2: Remember the quotient and write the remainder as the last digit in the answer (if no remainder, write 0).
Step 3: Divide the quotient from Step 2 by the desired base. Write the remainder as the nexttolast digit of the answer.
Step 4: Continue this process until the quotient and remainder are 0.
Let’s see some examples.
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[*16.3 *]Base a to Base a^n
There comes a special property of numbers when you have to convert from, say base 2 to base 4 or base 8, base 3 to base 9, etc. The idea is grouping digits.
This is best explained with an example. Let’s say you want to solve the following question:
Since 4 is the second power of 2, we will be taking groups of two (the power). For each grouping, starting from the right and working left, convert the number to base 4. This can be done by using the place value idea in 16.1 to convert to base 10, and then to the desired base (Note: when converting to bases under 10, you can imagine converting to base 10 as the number will be the same). For our example, this will be a digit between 0 and 3. This becomes the last digit of the number in base 4. We continue to convert these groupings to convert the number.
So, we’ll start with 10110. We can see that. This means that our last digit is 2. (Note: we can just convert the number to base 10 since 4<10).
Next, we use 10110. Converting to base 4, we get. The nexttolast digit is a 1.
Finally, we’re left with a 1. Even though this is only one digit of the grouping, we can imagine a 0 in front of the number to make it a grouping. Converting to base 4, we get. The next digit is a 1.
Our final answer: Now let’s work some more examples.
Since 4 is the second power of 2, we’ll make groups of two again.
Now, we convert every grouping to base 4 to get a digit from 0 to 3.
Since 8 is the third power of 2, we’ll now make groups of 3.
Now, we convert every grouping to base 8 to get a digit from 0 to 7.
Since 16 is the fourth power of 2, we’ll now make groups of 4.
Now, we convert every grouping to base 16 to get a digit from 0 to 9 or a letter from A to F.
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[*16.4 *]Decimals in Other Bases
We can use place values to convert decimals from a different base to base 10 just like how we used it to convert integers in Trick 16.1. For example:
However, calculating this value for any number besides base 10 can get quite lengthy. Instead, there’s a little trick to convert decimals in other bases:
Step 1: Imagine the number doesn’t have a decimal. Convert the integer to base 10. This is the numerator of the base 10 fraction.
Step 2: Raise the base to the power of the number of digits there are in the original decimal. This is the denominator of the base 10 fraction.
Let’s see some examples.
You can also reverse the process, going from a fraction or decimal in base 10 to a decimal in another base. In most cases, the denominator of the fraction in base 10 is a power of the base itself; if this isn’t the case, then a multiple of the denominator will be a power. After getting the denominator to be a power of the base, you simply have to convert the numerator of the base 10 fraction to the desired base. Then, just add the decimal in the right position based on the power of the base. Voila!
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[*17.1 *]Solving Linear Equations
Not every question has a trick on a Number Sense test. The majority of such questions pertain to algebra. There really is no foolproof trick for solving equations besides working it out. However, there are some pointers I can provide to make the process a bit easier.
Always look at what’s asked for. On certain occasions what’s being asked may simplify the calculations you need to do. For example:
In this case, rather than trying to solve for and then plug into the second expression, we can simply subtract 11 from both sides to get the answer. Awareness is key.
Simplify your calculations as much as possible. Cancel out like terms and bring the variables to one side before dividing. Keep the variable coefficient positive if possible. Little things like these make problems with equations much easier.
Practice. Above all, practice with a variety of questions will make you go faster. Here are some problems you can work on.
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[*17.2 *]Slope and yintercept
There are many ways to represent a line with an equation; pointslope, standard, slopeintercept, etc. However, the only one that will tell us both the slope and the yintercept is the slopeintercept form, or:
In Number Sense, most every questions asks about the slope, giving you either an equation or two points. Let’s look at each case.
When given an equation, the goal is to get it into slopeintercept form. However, instead of converting the entire equation to the desired form, we only need to focus on the value of the slope. That said, we can ignore the constant value in the original equation and simply need to move the term to be on the opposite side of the term and make the coefficient of to be 1.
Notice how we didn’t touch the 6. Since we knew we were looking for the slope, we ignored the constant value. If we were asked for the yintercept, we would follow similar steps except ignore the term and focus on the constant term.
The other way Number Sense asks questions about slope is with two coordinates. Since slope is, we can use the coordinates given to directly calculate the slope.
If asked for the yintercept given two points, first find the slope as shown above. Then, choose any of the two points (preferably the one with positive, small numbers) and plug into this equation:
From there, you would distribute the lefthand side and then add the constant from the righthand side. This would transform the equation into slopeintercept, from which you can find the yintercept.
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[*17.3 *]Inequalities
In recent years they have been asking more and more questions about inequalities, including the number of and the sum of all integer solutions. Since the derivations for each of these tricks are quite long, I’ll simply list the formulas. Just a note, means that you round to the greatest integer less than it, aka remove the decimal. Conversely, means you round to the least integer greater than it (round up).
[*18.1 *]Roots
A quadratic equation takes this form:
In this equation,,, and are constants (can be 0). Notice how the highest degree in the equation is 2.
To determine the solutions of this equation, there are a variety of formulas. The most prominent one is the Quadratic Formula, a lengthy yet foolproof way to determine the real or imaginary roots (solutions) of any quadratic formula. However, since the formula is complicated and timeconsuming and because Number Sense only asks questions with rational solutions, we won’t use this.
Another viable method is Completing the Square, which many people are taught in algebra and find to be fast and simple. However, the problem is that Number Sense usually asks questions with a value other than 1 for, making this trick significantly more difficult to do quickly. For this reason, we’ll skip it.
So now, what do we do? Well, one thing to note is that every time you are given quadratic given on a Number Sense test and asked to find a root, the quadratic will be factorable. The answer can’t be irrational. Thus, we will work to factor the quadratic directly to find the roots.
[*Step 1: *]Multiply
[*Step 2: *]Find two numbers that add to and multiply to the product found in Step 1. This step is the hardest part of this process, and practice is the only thing that’ll make it faster.
Step 3: Split up the term such that the two numbers found in Step 2 are the coefficients.
Step 4: Factor out the common term from each pair of terms.
Step 5: Combine.
Step 6: Solve each binary term to find the factors.
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[*18.2 *] Vieta’s Formulas
Vieta’s formula is a simple way to determine the sum and the product of roots of a quadratic equation. Rather than having to factor out quadratics to determine their sum, we can use this failproof formula to determine the value we need.
Some things to note:
1) Sum of 3 roots taken 2 at a time is
2) The sum of roots is always negative. Then the signs alternate.
[*19.1 *]Subsets
A set is simply a collection of different things. These things can be numbers, characters, symbols, etc. It’s denoted with curly brackets around the elements and commas in between.
A subset is any portion of a set; every member of the subset can be found in the set. Thus, for a set of elements, a subset can have 0 to elements in it.
To count the number of possible subsets of a set of elements, use the following formula:
This works because every object in the set can either be in the subset or not, giving each item 2 options.
[*19.2 *]Proper and Improper Subsets
An improper subset is the subset that contains every member of the set; that is, the subset and the set are identical. For example, given the set, the subset would be improper.
Every other possible subset, from 0 objects to objects, is a proper subset. Continuing with the example above, the subsets,,,,,, and are all proper.
For every set, there can only be 1 improper subset. That means every set has proper subsets. With this, you can solve some practice problems.
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[*19.3 *]Intersection and Union of Sets
The intersection and union of sets is very similar to GCD and LCM, discussed in Chapter 3.
The intersection of two sets are the elements found in both sets. The intersection of two sets, set A and set B, is denoted A∩B. It is analogous to the GCD of numbers.
The union of two sets contains every element found in both sets, but only once, as a set cannot repeat an element. The intersection of two sets, set A and set B, is denoted A∪B. It is analogous to the LCM of numbers.
Another property that you should know about sets is that:
The number of elements in A∩B and A∪B is equal to the number of elements in each of the sets combined.
Here are some examples.
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[*19.4 *]Cartesian Product
The Cartesian product of two sets is basically the set of all ordered pairs using one element from each set. For example, if you have one set be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} and another set be {spades, clubs, hearts, diamonds}, then the Cartesian Product will result in the 52 cards in a deck. To determine the number of ordered pairs in the Cartesian product of two sets, use this formula:
That’s it. Just multiply the number of elements in each set to get the number of ordered pairs in the Cartesian product.
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[*20.1 *]Solving Equations with Logarithms
Just how subtraction is the inverse operation of addition, a logarithm is the inverse operation for an exponent. An exponent raises a number to a power: for example. A logarithm works by asking, “To what power do I need to raise 3 to get 9?”
The notation works as shown:
Some key things to know are:
In Number Sense, questions with logarithms usually give you the exponent and the base, asking for a variable in the power. The trick is to raise the base to the exponent given, and then set the value equal to the expression found in the power. Let’s see some examples.
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[*20.2 *]Nested Logarithms
Nested logarithms are basically logarithms inside one another. They look like this:
The key when working with any nested functions is to start from the inside and work your way out. With Number Sense, every logarithm will yield a rational number, so the answer to the insidemost logarithm will likely be an integer power of the next base. Let’s see each step for the example above.
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[*21.1 *]Adding and Subtracting Matrices
A matrix is basically an array of numbers with rows and columns. Each cell can contain a number, symbol, or other character. For the purpose of Number Sense, matrices will contain numbers. It will look kinda like this:
The matrix above is a 2×3 matrix because there are 2 rows and 3 columns.
When adding and subtracting matrices, first you need to make sure that all matrices have the same dimensions. A 2×3 matrix is not the same as a 3×2 matrix. Then, you simply add or subtract each corresponding cell to get the final answer. Also, when you multiply a number by a matrix, you simply multiply every term in the matrix by that number. Let’s look at some examples.
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[*21.2 *]Multiplying Matrices
To be able to multiply two matrices, first the dimensions of the matrices must be compatible. The number of columns of the first matrix must equal the number of rows of the second matrix. This is the only condition that needs to be met when multiplying two matrices with numbers in their cells.
Now, since every Number Sense question has to have an answer, the dimensions of the matrices will be compatible. To multiply two matrices, follow these steps:
[*Step 1: *]Make sure the number of columns of the first matrix is equal to the number of rows of the second matrix.
Step 2: Find the dot product of row 1 of the first matrix and column 1 of the second matrix to get the term in row 1 and column 1 of the answer. In a dot product, you multiply the first term in the row with the first term in the column, the second with the second, and so on until every number has been used. You then add every product.
Step 3: Find the dot product of row 2 of the first matrix and column 1 of the second matrix to get the term in row 2 and column 1 of the answer. Continue to change the row number and column number to get the values for the cells in the resultant matrix.
The steps may seem confusing without an example, so let’s change that.
Just a quick note, most Number Sense problems will show up in the form of the last example as the others not only don’t yield an single number as an answer. However, Number Sense may ask for any one term of the matrix, for example row 2 column 2. If this is asked, simply do the calculations for that cell and ignore the rest.
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[*21.3 *]Determinants
The determinant of a matrix is a number that is used in a number of calculations in calculus and other higherlevel math. The calculation varies for each dimension, but most questions ask about 2×2 matrices. The formula for the determinant is:
Note that the vertical bars around a matrix means the determinant. As for a 3×3 matrix, we use this formula:
If you look at each of the terms in the 2×2 determinants, you can see a pattern. Now let’s see some examples.
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1) 8080 2) 7914 3) 1443 4) 1086
5) 117 6) 1002 7) 2231 8) 1370
9) 1011 10) 1315 11) 7285 12) 2627
13) 651 14) 33443 15) 1990 16) 2278
17) 297 18) 788 19) 2100 20) 3149
1) 502 to 554 2) 2831 to 3129 3) 10507 to 11613
4) 43773 to 48381 5) 4109 to 4541 6) 4267 to 4717
7) 1602 to 1770 8) 4677 to 5169 9) 1279 to 1413
10) 22038 to 24358 11) 7493 to 8281 12) 47575 to 52583
13) 2774 to 3066 14) 12824 to 14174
1) 484 2) 385 3) 803 4) 539
5) 946 6) 2057 7) 2530 8) 429
9) 8679 10) 2838 11) 52725 12) 23754
13) 46398 14) 40959 15) 109557 16) 4995
17) 123321 18) 2553 19) 3872 20) 1694
1) 600 2) 3 3) 1825 4) 925
5) 9075 6) 1150 7) 320 8) 3425
9) 910 10) 3.1 11) 631.25 12) 300.4
1) 3000 2) 15 3) 9125 4) 4625
5) 212.5 6) 18000 7) 1600 8) 17125
9) 4550 10) 15.5 11) 3156.25 12) 1502
1) 540 2) 560 3) 2100 4) 2730
5) 1150 6) 8320 7) 35505 8) 1330
9) 2430 10) 1890 11) 3600 12) 3910
13) 3960 14) 9610 15) 10260 16) 3610
17) 2990 18) 7290 19) 9000 20) 11560
1) 707 2) 1515 3) 2323 4) 12423
5) 26058 6) 2424 7) 7007 8) 14014
9) 135135 10) 1235234 11) 36948912 12) 4913953
1) 10608 2) 10815 3) 11536 4) 12168
5) 92412 6) 12198 7) 12995 8) 13689
9) 22898 10) 96368 11) 499849 12) 45156
13) 46398 14) 835347
1) 9506 2) 9118 3) 8736 4) 8463
5) 9024 6) 8188 7) 8645 8) 7832
9) 8455 10) 2838 11) 7656 12) 8178
1) 2021 2) 1085 3) 399 4) 624
5) 8099 6) 1591 7) 2484 8) 1224
9) 10197 10) 396 11)2091 12) 23754
13) 2688 14) 891
1) 216 2) 7224 3) 5609 4) 1221
5) 946 6) 624 7) 11009 8) 27216
9) 13221 10) 30616
1) 616 2) 2812 3) 2408 4) 2988
5) 3312 6) 2848 7) 6688 8) 2788
9) 3871 10) 2726 11) 7566 12) 2988
13) 3172 14) 14012 15) 16263 16) 16244
17) 66465 18) 168644 19) 390486 20) 202768
1) 625 2) 1225 3) 2025 4) 3025
5) 4225 6) 5625 7) 7225 8) 9025
9) 11025 10) 13225 11) 15625 12) 18225
1) 810 2) 160 3) 1690 4) 2890
5) 405 6) 8410 7) 2450 8) 909
1) 421 2) 613 3) 841 4) 1301
5) 2521 6) 11101 7) 3961 8) 5101
9) 313 10) 7081 11) 3121 12) 26221
1) 1717 2) 5050 3) 808 4) 4040
5) 4545 6) 6565 7) 3232 8) 8989
9) 1010 10) 8585 11) 9090 12) 9797
1) 86569 to 95681 2) 308 to 340 3) 2770 to 3062
4) 616 to 680 5) 13042 to 14414 6) 86569 to 95681
7) 27133 to 29989 8) 821 to 907 9) 1584 to 1751
10) 315187 to 348365 11) 188 to 207 12) 2083692 to 2303028
1) 4 2) 9 3) 6 4) 18
5) 1 6) 17 7) 2 8) 6
9) 49 10) 4 11) 13 12) 7
13) 6 14) 24
1) 48 2) 1080 3) 180 4) 180
5) 450 6) 476 7) 800 8) 168
9) 686 10) 144 11) 840 12) 672
13) 65 14) 70
1) 6 2) 2 3) 0 4) 5
5) 6 6) 1 7) 3 8) 3
9) 4 10) 0 11) 5 12) 5
13) 5 14) 6 15) 7 16) 3
7) 2 18) 4 19) 4 20) 4
1) 2 2) 0 3) 0 4) 3
5) 3 6) 0 7) 1 8) 3
9) 7 10) 1
1) 12 2) 1 3) 1 4) 1
5) 1 6) 1 7) 0 8) 1
9) 8 10) 1
1) 37 2) 169 3) 127 4) 217
5) 397 6) 271 7) 817 8) 631
9) 2611 10) 4921
1) 3 2) 4 3) 56 4) 2
5) 512 6) 3125 7) 288 8) 16
1) 6 2) 2 3) 5 4) 3
5) 3 6) 285 7)
1) 9 2) 3) 125 4)
5) 49 6) 1024 7) 8) 2197
9) 27 10) 27
1) 3421 to 3781 2) 115 to 127 3) 217 to 240
4) 244 to 269 5) 1160 to 1282 6) 1621 to 1792
7) 3406 to 3765 8) 7554 to 8349 9) 1259 to 1391
10) 279 to 309 11) 487 to 538 12) 559 to 618
1) 6 2) 6 3) 8 4) 4
5) 6 6) 2 7) 6 8) 12
9) 15 10) 8
1) 28 2) 31 3) 42 4) 72
5) 91 6) 48 7) 266
1) 12 2) 16 3) 42 4) 40
5) 16 6) 24 7) 16
1) 12 2)  3) 4 4)
5) 6) 7)  8) 
1) 6 2) 4 3) 5 4) 9
5) 0 6) 10 7) 11 8) 9
1) 111 2) 1017 3) 555 4) 123
5) 1710 6) 2017 7) 1900 8) 209
9) 999 10) 3098 11) XVII 12) XXXV
13) MXXXVII 14) DCCLXV 15) CMLXXXVII 16) CXIV
17) CDXXXII 18) DXCVI 19) DCXCVIII 20) CMXCIX
1) 17 2) 27 3) 12 4) 218
5) 37 6) 42 7) 48 8) 16
9) 98 10) 101
1) 28 2) 75 3) 112 4) 92
5) 96 6) 24 7) 91 8) 78
9) 325 10) 27
1) 13 2) 15 3) 5 4) 469
5) 23 6) 125 7) 10 8) 26
1) 12 2) 5040 3) 30 4) 60
5) 840 6) 504 7) 11880 8) 1680
9) 1814400 10) 90
1) 6 2) 1 3) 15 4) 10
5) 35 6) 84 7) 495 8) 70
9) 45 10) 45
1) 276 2) 435 3)196 4) 42
5) 81 6) 300 7) 210 8) 266
9)140 10) 294 11) 203 12) 400
13) 220 14) 5050
1) 210 2) 240 3) 374 4) 375
5) 440 6) 297 7) 630 8) 430
9) 559 10) 138 11) 288 12) 1133
1) 256 2) 2 3) 4)
5) 6) 25
1) 88 2) 608 3) 517 4) 372
5) 282 6) 550 7) 520
1) 58163 to 64286 2) 1160 to 1282 3) 633 to 700
4) 162 to 179 5) 101 to 111 6) 372 to 411
7) 1402 to 1550 8) 56729 to 62700 9) 69 to 76
10) 17098 to 18898 11) 2399 to 2651 12) 35625 to 39375
1) 2) 3) 4) 3
5) 6) 7) 8)
9) 10)
1) 2) 3) 4)
5) 6) 1 7) 8)
1) 2) 3) 4)
5) 6) 7) 8)
9) 10)
1) 1 2) 3) 4)
5) 6) 7) 8)
9) 15 10)
1) 2)30 3) 4) 12
5) 42 6) 90 7) 8)
9) 10) 11) 12)
1) 2) 3) 4)
5) 6) 7) 8)
9) 10) 149 11) 12)
1) 2) 3) 6 4)
5) 5 6) 28 7) 8)
9) 10)
1) 2) 5 3) 7 4)
5) 6 6) 7) 8)
9) 10)
1) 2 2) 3 3) 2 4)
5) 6)4 7) 12 8)
1) 2) 3) 4)
5) 6) 7) 8)
9) – 10) 
1) 2) 3) 4)
5) 6) 7) 7777 8) 1313
9) 1461 10) 2162
1) 2) 3) 4)
5) 6) 7) 8)
9) 10) 1 11) 12) 5
13) 0777 14) 2555 15) 1444 16) 4566
17) 2666 18) 1272
1) 18 2) 40 3) 23 4) 63
5) 39 6) 199 7) 53 8) 412
9) 194 10) 416 11) 556 12) 1718
13) 2126 14) 636
1) 120 2) 54 3) 20 4) 33
5) 363 6) 25A 7) 132 8) 1253
9) 4D2 10) 6AB 11) 2404 12) 618
13) 1011011001 14) 152
1) 31 2) 52 3) 123 4) 235
5) 355 6) 33 7) 88 8) 111011010
9) 39B 10) 101010111100 11) 211210
12) 110100010 13) 111111 14) 121312
1) 2) 3) 4)
5) 6) .201 7) .22 8) .1A
9) .527 10) .88
1) 12 2) 14 3) 28 4) 8
5) 23 6) 1 7) 72 8)
1) 2) 3) 1 4) 
5) 5 6) 7) 8)
9) 10)
1) 2 2) 3) 4)
5) 6) 4 7) 1 8) 2 9)
1) 64 2) 16 3) 15 4) 128
5) 1 6) 511 7) 63
1) 4 2) 6 3) 2 4) 4
5) 8 6) 12 7) 2
1) 14 2) 30 3) 30 4) 24
5) 24 6) 24
1) 2 2) 2 3) 4) 8
5) 6) 2 7) 8 8) 8
9) 2 10) 10 11) 5 12) 27
1) 2) 1 3) 1 4) 2
5) 0 6) 2 7) 1 8) 3
1) 17 2) 98 3) 3 4) 4
5) 8 6) 27
1) 20 2) 31 3) 11 4) 13
5) 1 6) 120
1) 13 2) 249 3) 7 4) 180
5) 7 6) 2 7) 190
About this Book From those who read a poster asking them to join the school’s Number Sense team to the seasoned competitor eyeing State medals, this book serves as a guide and as inspiration, to not only pique interest but to build a foundation on. You’ll find in this book a stepbystep guide on howto solve Number Sense problems and prepare you for any mental math competitions. There are over 80 mental math techniques are explained in this book, and over 750 problems to learn, train, and improve with. Just a note: as fun as it is to learn new tricks and techniques, it’s just as important to master them and implement them in fulllength tests! I wish you all the best of luck, and happy computing!