Gmat Math 01 - Number Properties

Front 1

Increment

Back 1

These are sequences of numbers whose values go up or down by the same amount (the increment) from one item in the sequence to the next. For instance, the set {4, 7, 10, 13, 16} is evenly spaced because each value increases by 3 over the previous value.

Front 2

For all evenly spaced sets, just remember: the average equals = ?

Back 2

(First + Last) / 2. For example, What is the arithmetic mean of 4,8,12,16,20, and 24?
(24 + 4) / 2 = 14.

Front 3

The sum of the elements in the set equals?

Back 3

The arithmetic mean (average) number in the set times the number of items in the set.
For example, What is the sum of 4,8,12,16,20, and 24? The average is equal to 14. There are 6 terms, so the sum equals 14 x 6 = 84.

Front 4

How many multiples of 7 are there between 100 and 150?

Back 4

Here it may be easiest to list the multiples: 105, 112, 119, 126, 133, 140, 147. Count the number of terms to get the answer: 7. Alternatively, we could note that 105 is the first number, 147 is the last number, and 7 is the increment:
Number of terms = (Last - First) + Increment + 1 = (147 - 105) + 7 + 1 = 6 + 1 = 7.

Front 5

Odd numbers

Back 5

Нечётные числа

Front 6

The average of an odd number of consecutive integers (1, 2, 3, 4, 5) will always be?

Back 6

An integer. This is because the "middle number" will be a single integer.

Front 7

The average of an even number of consecutive integers (1, 2, 3, 4) will never be?

Back 7

An integer (2.5), because there is no true "middle number.

Front 8

According to the Factor Foundation Rule...

Back 8

According to the Factor Foundation Rule, every number is divisible by all the factors of its factors.

Front 9

For any set of consecutive integers with an ODD
number of items, the sum of all the integers is ALWAYS a multiple of the number of items. Why?

Back 9

This is because the sum equals the average times the number of items. For an odd number of integers, the average is an integer, so the sum is a multiple of the number of items. The average of {13, 14, 15, 16, 17} is 15, so 15 x 5 = 13 + 14 + 15 + 16 + 17.

Front 10

For any set of consecutive integers with an EVEN number of items, the sum of all the items is NEVER a multiple of the number of items. Why?

Back 10

This is because the sum equals the average times the number of items. For an even number of integers, the average is never an integer, so the sum is never a multiple of the number of items. The average of {8, 9, 10, II} is 9.5, so 9.5 x 4 = 8 + 9 + 10 + 11. That is, 8 + 9 + 10 + 11 is NOT a multiple of 4.

Front 11

If x is an even integer, is x(x + 1)(x + 2) divisible by 4?

Back 11

x(x + 1)(x + 2) is the product of 3 consecutive integers, because x is an integer. If there is one even integer in a series of consecutive integers, the product of the series is divisible by 2. If there are two even integers in a series of consecutive integers, the product of the series is divisible by 4. Set up prime boxes:
If x is even then x + 2 is even, so 2 is a factor of
x(x + l)(x + 2) twice. Therefore, the product 2 x 2 = 4 is a
factor of the product of the series. The answer to the question given above is "Yes."

Front 12

The result of dividing two integers is sometimes an integer. The result is called ...

Back 12

The quotient

Front 13

Factor

Back 13

Divisor. Множитель.
2 and 5 are factors of 10. — 2 и 5 являются множителями 10. A factor is a positive integer that divides evenly into an integer. 1,2,4 and 8 are all the factors (also called divisors) of 8.

Front 14

An integer is divisible by 2 if...

Back 14

The integer is EVEN.

Front 15

An integer is divisible by 3 if...

Back 15

The SUM of the integer's DIGITS is divisible by 3.

Front 16

An integer is divisible by 4 if...

Back 16

The integer is divisible by 2 TWICE, or if the LAST lWO digits are divisible by 4.

Front 17

An integer is divisible by 6 if...

Back 17

The integer is divisible by BOTH 2 and 3.

Front 18

An integer is divisible by 8 if...

Back 18

The integer is divisible by 2 THREE TIMFS, or if the LAST THREE digits are divisible by 8.

Front 19

An integer is divisible by 9 if...

Back 19

The SUM of the integer's DIGITS is divisible by 9.

Front 20

If you add or subtract multiples of N, the result is ...

Back 20

The result is a multiple of N. You can restate this principle using any of the disguises: for instance, if N is a divisor of x and of y, then N is a divisor of x +y.

Front 21

The number 1 is prime?

Back 21

Note that the number 1 is not considered prime, as it has only one factor (itself). Thus, the first prime number is 2, which is also the only even prime.

Front 22

The GMAT expects you to know the factor foundation rule. What is this rule?

Back 22

If a is a factor of b, and b is a factor of c, then a is a factor of c.

Front 23

GCF

Back 23

Greatest Common Factor. The largest divisor of two or more integers .

Front 24

LCM

Back 24

Least Common Multiple. The smallest multiple of two or more integers.

Front 25

Note that if two numbers have NO primes in common, then their GCF is ... and LCM is ...

Back 25

Note that if two numbers have NO primes in common, then their GCF is 1 and their LCM is simply their product.

Front 26

A multiple

Back 26

Кратное число. 12 is a multiple of 3 двенадцать кратно трём.

Front 27

Consecutive integers alternate between?

Back 27

Even and odd

Front 28

Odd +- Even =

Back 28

Odd

Front 29

Odd +- Odd =

Back 29

Even

Front 30

Even +- Even

Back 30

Even

Front 31

Odd x Odd

Back 31

Odd

Front 32

Even x Even

Back 32

Even (and divisible by 4)

Front 33

Odd x Even

Back 33

Even

Front 34

If there are TWO even integers in a set of integers being multipled together, the result will be divisible by?

Back 34

4

Front 35

If there are THREE even integers in a set of integers being multipled together, the result will be divisible by?

Back 35

8

Front 36

An odd number divided by any other integer CANNOT produce

Back 36

An even integer

Front 37

An odd number divided by an even number CANNOT produce

Back 37

An integer

Front 38

If you see a sum of two primes that is odd, one of those primes must be the number...

Back 38

2

Front 39

If you know that 2 CANNOT be one of the primes in the sum, then the sum of the two primes must be...

Back 39

Even

Front 40

If x > 1, what is the value of integer x?

(1) There are x unique factors of x.
(2) The sum of x and any prime number larger than x is odd.

Back 40

Statement (1) tells us that there are x unique factors of x. In order for this to be true, EVERY integer between 1 and x, inclusive, must be a factor of x. Testing numbers, we can see that this property holds for 1 and for 2, but not for 3 or for 4. In fact, this property does not hold for any higher integer, because no integer x above 2 is divisible by x-1. Therefore,
x = 1 or 2. However, the original problem stem told us that x > 1, so x must equal 2. SUFFICIENT.
Statement (2) tells us that x plus any prime number larger than x is odd. Since x > 1, x must equal at least 2, so this includes only prime numbers larger than 2. Therefore, the prime number is odd, and x is even. However, this does not tell us which even number x could be. INSUFFICIENT.

Front 41

When is |x - 4| equal to 4 - x?

Back 41

5. x <=4. Absolute value brackets can only do one of two things to the expression inside of them: (a) leave
the expression unchanged, whenever the expression is 0 or positive, or (b) change the sign of the whole
expression, whenever the expression is 0 or negative. (Notice that both outcomes occur when the expression
is zero, because "negative 0" and "positive 0" are equal.) In this case, the sign of the whole expression
x - 4 is being changed, resulting in -(x - 4) = 4 - x. This will happen only if the expression x - 4 is 0 or
negative. Therefore x - 4=0, or x<=4.

Front 42

Exponent

Back 42

Экспонента, показатель степени

Front 43

Numbers can be either positive or negative except...

Back 43

The number 0, which is neither

Front 44

The absolute value of a number answers which question

Back 44

How far away is the number from 0 on the number line? For example, the number 5 is exactly 5 units away from 0, so the absolute value of 5 equals 5. Mathematically, we write this using the symbol for absolute value: |5|= 5. To find the absolute value of -5, look at the number line above: -5 is also exactly 5 units away from 0. Thus, the absolute value of -5 equals 5, or, in mathematical symbols, |-5| = 5.

Front 45

When you multiply or divide a group of nonzero
numbers, the result will be positive if ...

Back 45

You have an EVEN number of negative numbers. The
result will be negative if you have an ODD number of negative numbers.

Front 46

Is the product of all of the elements in Set S negative?
(1) All of the elements in Set S are negative.
(2) There are 5 negative numbers in Set S.

Back 46

This is a tricky problem. Based on what we have learned so far, it would seem that Statement (2) tells us that the product must be negative. (5 is an odd number, and when the GMAT says "there are 5" of something, you CAN conclude there are EXACTLY 5 of that thing).
However, if any of the elements in Set 5 equals zero, then the product of the elements in Set 5 will be zero, which is NOT negative. Therefore Statement (2) is INSUFFICIENT.
Statement (1) tells us that all of the numbers in the set are negative. If there are an even number of negatives in Set 5, the product of its elements will be positive; if there are an odd number of negatives, the product will be negative. This also is INSUFFICIENT.
Combined, we know that Set 5contains 5 negative numbers and nothing else. SUFFICIENT. The product of the elements in Set 5 must be negative. The correct answer is (C).

Front 47

How to read 4^3?

Back 47

The expression is read as "four to the third power." It consists of a base (4) and an exponent (3). It may be called as cube too.

Front 48

THE EVEN EXPONENT IS DANGEROUS. WHY?

Back 48

IT HIDES THE SIGN OF THE BASE!

Front 49

x^6 = x^7 = x^15. What is x?

Back 49

x must be either 0 or 1

Front 50

When the base of an exponential expression is a positive proper fraction (in other words, a fraction
between 0 and 1), what occurs?

Back 50

As the exponent increases, the value of the expression decreases!

Front 51

ADDING EXPONENTS

Back 51

When multiplying two terms with the same base, combine exponents by adding.

Front 52

SUBTRACTING EXPONENTS

Back 52

When dividing two terms with the same base, combine exponents by subtracting.

Front 53

NESTED EXPONENTS

Back 53

When raising a power to a power, combine exponents by multiplying.

Front 54

AN EXPONENT OF 0

Back 54

Any nonzero base raised to the power of zero (e.g. 3^0) is equal to 1.

Front 55

What is 25^(3/2)

Back 55

3 - numerator, 2 - denominator

Front 56

If r^3 + /r/= 0, what are the possible values of r?

Back 56

15.0,-1: If r^3 + /r/ = 0, then r^3 must be the opposite of /r/. The only values for which this would be true are 0, which is the opposite of itself, and -1, whose opposite is 1.

Front 57

A root, also called ...?

Back 57

Radical and it is the opposite of an exponent, in a sense.

Front 58

Even roots only have a which value?

Back 58

Positive

Front 59

For odd roots (cube root, 5th root, 7th root, etc.), the root will have ... as the base.

Back 59

The same sign

Front 60

WHEN CAN YOU SIMPLIFY ROOTS?

Back 60

You can only simplify roots by combining or separating them in multiplication and division.
You cannot combine or separate roots in addition or subtraction.

Front 61

PEMDAS

Back 61

The correct order of operations is: Parentheses -Exponents - (Multiplication - Division) - (Addition - Subtraction)

Front 62

If P is an integer, is P/18 an integer?

(1) 5p/18 is an integer.
(2) 6p/18 is an Integer.

Back 62

Statement (1) can be restated as 5p is divisible by 18, or by two 3's and a 2. A double prime box, as shown to the right, allows you to separate the factors of a larger product. The coefficient of 5 does not provide ANY of the necessary primes to be divisible by 18 (in other words, it does not have any 3's or 2's as factors). Therefore, in order for 5p to be divisible by 18, P must be divisible by two 3's and a 2. Thus, statement (1) can be rephrased: There are (at least) two 3's and a 2.in the prime box of p. This is sufficient to answer the question.
Statement (2) can be restated as 6p is divisible by 18 or by two 3's and a 2. A double prime box, as shown to the right, allows you to separate the factors of a larger product. The coefficient of 6 provides two of the necessary primes to be divisible by 18 (6 has a factor 00 and a factor of 2). Therefore, in order for 6p to be divisible by 18, P must be divisible by one 3. While there may be additional primes in the prime box of p, statement (2) only guarantees one 3 in p's prime box. Thus, statement (2) can be rephrased: There is (at least) one 3 in the prime box of p. This is not sufficient to answer the question.

Front 63

VALUE: These questions require you to solve for one numerical value.

Back 63

The information in the statements can be considered sufficient if it allows you to find a single number to answer the question. If the information yields more than one value, the information is insufficient.

Front 64

YES/NO: These questions require you to give a simple yes or no answer.

Back 64

The information in the statements can be considered sufficient if it allows you to conclusively answer YES or NO. If the answer is MAYBE, the information is insufficient.

Front 65

List all the primes up to 100

Back 65

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Total 25 numbers.

Front 66

Some additional facts about primes that may be helpful on the GMAT.

Back 66

(1) There is an infinite number of prime numbers.
(2) There is no simple pattem in the prime numbers.
(3) Positive integers with only 2 factors must be prime, and positive integers with more than two factors are never prime.

Front 67

If you add or subtract multiples of an integer, you get another multiple of that integer. We can now generalize to other situations.

Back 67

(1) If you add a multiple of N to a non-multiple of N, the result is a non-multiple of N. (The same holds true for subtraction). 18 - 10 = 8 (Multiple of 3) - (Non-multiple of 3) = (Non-multiple of 3)
(2) If you add two non-multiples of N, the result could be either a multiple of N or a non-multiple of N.
19 + 13 = 32 (Non-multiple of 3) + (Non-multiple of 3) = (Non-multiple of 3). 19 + 14 = 33 (Non-multiple of 3) + (Non-multiple of 3) = (Multiple of 3)

Front 68

Three general properties of the GCF and LCM are worth noting:

Back 68

(1) (GCF of m and n) * (LCM of m and n) = m * n.
(2) The GCF of m and n cannot be larger than the difference between m and n.
(3) Consecutive multiples of n have a GCF of n. For example, 8 and 12 are consecutive multiples of 4. Thus 4 is a common factor of both numbers. But 8 and 12 are exactly 4 units apart. Thus 4 is the greatest possible common factor of 8 and 12. (For this reason, the GCF of any two consecutive integers is 1, because both integers are multiples of 1 and the numbers are 1 unit apart).

Front 69

One special property of perfect squares is that all perfect squares have an ... number of total factors.
Similarly, any integer that has an ... number of total factors MUST be a perfect square.

Back 69

Odd

Front 70

The prime factorization of a perfect square contains only ... powers of primes. It is also true that any number whose prime factorization contains only ... powers of primes must be a perfect square.

Back 70

Even

Front 71

N! is a multiple of all the integers from 1 to N.

Back 71

Because it is the product of all the integers from 1 to N. Any factorial N! must be divisible by all integers from 1 to N.

Front 72

If you need a number that leaves remainder R upon division by N, simply take any multiple of N and add R to it. This is equivalent to writing ...

Back 72

x = Q x N + R, since Q x N is a multiple of N.

Front 73

If a / b yields a remainder of 5, c / d yields a remainder of 8, and a, b, c and d are all integers, what is the smallest possible value for b + d?

Back 73

Since the remainder must be smaller than the divisor, 5 must be smaller than b. b must be an integer, so b is at least 6. Similarly, 8 must be smaller than d, and d must be an integer, so d must be at least 9. Therefore the smallest possible value for b + dis 6 + 9 = 15.

Front 74

Two useful tips for arithmetic with remainders, if you have the same divisor throughout:

Back 74

(1) You can add and subtract remainders directly, as long as you correct excess or negative remainders.
(2) You can multiply remainders, as long as you correct excess remainders at the end.

Front 75

When positive integer A is divided by positive integer B, the result is 4.35. Which of the following could be the remainder when A is divided by 8?

(A) 13 (B) 14 (C) 15 (D) 16 (E) 17

Back 75

We isolate the decimal part of the division result: 0.35. Now we set that decimal equal to the unknown remainder R divided by the divisor B:
0.35 = Remainder/Divisor = R/B
Now, express the decimal as a fraction and reduce: 0.35 = 35/700 = 7/20 = R/B
Finally, we cross-multiply: 7B=20R
Now, since both B and R are integers, we can see that R must contain a 7 in its prime factorization;
otherwise, there is no way for a 7 to appear on the left side. Thus, R must be a multiple of 7. The only answer choice that in multiple of 7 is 14, which is the correct answer.

Front 76

Is the sum of integers a and b divisible by 7?

(1) a is not divisible by 7.
(2) a - b is divisible by 7.

Back 76

We understand that 1 and 2 are insufficient alone. Now let's consider the statements together.

I) says that a is not divisible by 7. Hence, lets take a=24, 28 or something. Now a-b is divisible by 7 which implies a-b will give a number divisible by 7. for e.g if a=24, then b could be 3, 10 or so, for a-b to be divisible by 7. Here if we note, all b does is remove the remainder obtained when a is divided by 7 from a. Since, a is not divisible by 7, a will be taking some value between x and x+7 where x is a random number. So a cannot be x and x+7. a could be x+1, x+2, x+3, x+4, x+5, x+6,. Based on b, b will have to remove this 1,2,3,4,5,6 from a (Note- it can also remove something like 10 from 24 which is 7+3 and hence a way to remove the extra 3 from a) Now , lets consider a+b. since b removes the extra number from 1 to make it divisible by 7, it can never add the complement of that factor to a to make it divisible by 7.

As an example, see below. It would be easier to understand the above explanation with this example.

For e.g, a=10, b=3 and its complement( The number required to be added to a to make it divisible by 7) would be c(b)=7-3 B cant be 3 and 4 at the same time and thus can never make a-b and a+b divisible by 7 at the same time.

II) (1) --> know nothing about b. Not sufficient.
(2) --> --> Nor sufficient

(1)+(2) and --> .

--> is divisible by 7. Since is remainder from a being divided by 7, and thus is not divisible by 7. So, is not divisible by 7. Sufficient.

Front 77

Which of the following numbers is NOT prime?

(A) 6! - 1 (8) 6! + 21 (C) 6! + 41 (0)7! - 1 (E) 7! + 11

Back 77

try to find an answer choice which CANNOT be-prime based on the
properties of divisibility. Earlier in this chapter, we learned the following property of factorials and divisibility:
N!is a multiple of all integers from 1 to N. We can apply this concept directly roche answer choices:
(A) 6! - 1: 6! is not prime, but 6! - ·1·might be prime, because 6! and 1 do not share any prime factors.
(B) 61+21: 61 is not prime, and. 61+ 21 CANNOT be prime, because 6!and 21 are both multiples
of 3. Therefore, 61+21 is divisible by 3.
(C) 6! + 41: 6! is not prime, but 6! + 41 might be prime, because 6! and 41 do not share any prime
factors.
(D) 7! - 1: 7! is not prime, but 7! - 1 might be prime, because 7! and Ldenot share anY'prime factors.
(E) 7! + 11: 7! is not prime, but 7! + 11 might be prime, because 7! and 11 do not share any prime
factors.

Front 78

Explain a general rule of thumb in factoring and distributing of exponents

Back 78

A general rule of thumb is that when you encounter any exponential expression in which two or more terms include something common in the base, you should consider factoring. Similarly, when an expression is given in factored form, conslder distributing it.

Front 79

Simplify 4 / (3 - 2^(1/2))

Back 79

To simplify this type of problem, we need to use the conjugate of the denominator. The conjugate for any square root expression involving addition or subtraction is defined as follows:

For a+(b^1/2), the conjugate is given by a-(b^1/2).
For a-(b^1/2), the conjugate is given by a+(b^1/2).