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79 Cards in this Set
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Increment

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.


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

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

The sum of the elements in the set equals?

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. 

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

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. 

Odd numbers

Нечётные числа


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

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


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

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


According to the Factor Foundation Rule...

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


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? 
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.


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?

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.


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

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." 

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

The quotient


Factor

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. 

An integer is divisible by 2 if...

The integer is EVEN.


An integer is divisible by 3 if...

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


An integer is divisible by 4 if...

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


An integer is divisible by 6 if...

The integer is divisible by BOTH 2 and 3.


An integer is divisible by 8 if...

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


An integer is divisible by 9 if...

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


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

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.


The number 1 is prime?

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.


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

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


GCF

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


LCM

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


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

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


A multiple

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


Consecutive integers alternate between?

Even and odd


Odd + Even =

Odd


Odd + Odd =

Even


Even + Even

Even


Odd x Odd

Odd


Even x Even

Even (and divisible by 4)


Odd x Even

Even


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

4


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

8


An odd number divided by any other integer CANNOT produce

An even integer


An odd number divided by an even number CANNOT produce

An integer


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

2


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

Even


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. 
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 x1. 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. 

When is x  4 equal to 4  x?

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. 

Exponent

Экспонента, показатель степени


Numbers can be either positive or negative except...

The number 0, which is neither


The absolute value of a number answers which question

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.


When you multiply or divide a group of nonzero
numbers, the result will be positive if ... 
You have an EVEN number of negative numbers. The
result will be negative if you have an ODD number of negative numbers. 

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. 
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). 

How to read 4^3?

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.


THE EVEN EXPONENT IS DANGEROUS. WHY?

IT HIDES THE SIGN OF THE BASE!


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

x must be either 0 or 1


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


ADDING EXPONENTS

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


SUBTRACTING EXPONENTS

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


NESTED EXPONENTS

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


AN EXPONENT OF 0

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


What is 25^(3/2)

3  numerator, 2  denominator


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

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.


A root, also called ...?

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


Even roots only have a which value?

Positive


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

The same sign


WHEN CAN YOU SIMPLIFY ROOTS?

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

PEMDAS

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


If P is an integer, is P/18 an integer?
(1) 5p/18 is an integer. (2) 6p/18 is an Integer. 
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. 

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

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.


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

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.


List all the primes up to 100

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.


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

(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. 

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

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

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

(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). 

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. 
Odd


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.

Even


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

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.


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 ...

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


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?

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.


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

(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. 

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 
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 crossmultiply: 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. 

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. 
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 ab is divisible by 7 which implies ab will give a number divisible by 7. for e.g if a=24, then b could be 3, 10 or so, for ab 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)=73 B cant be 3 and 4 at the same time and thus can never make ab 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. 

Which of the following numbers is NOT prime?
(A) 6!  1 (8) 6! + 21 (C) 6! + 41 (0)7!  1 (E) 7! + 11 
try to find an answer choice which CANNOT beprime 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. 

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

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.


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

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). 