GRE_Math_Bible_eBook

42GRE Math Bible

17.If each of the three nonzero numbers a, b, and c is divisible by 3, then abc must be divisible by which one of the following the numbers?

(A)8

(B)27

(C)81

(D)121

(E)159

18.How many positive five-digit numbers can be formed with the digits 0, 3, and 5?

(A)14

(B)15

(C)108

(D)162

(E)243

19.If n is a positive integer, which one of the following numbers must have a remainder of 3 when divided by any of the numbers 4, 5, and 6?

(A)12n + 3

(B)24n + 3

(C)80n + 3

(D)90n + 2

(E)120n + 3

21.The number 3072 is divisible by both 6 and 8. Which one of the following is the first integer larger than 3072 that is also divisible by both 6 and 8?

(A)3078

(B)3084

(C)3086

(D)3090

(E)3096

22.If the least common multiple of m and n is 24, then what is the first integer larger than 3070 that is divisible by both m and n?

(A)3072

(B)3078

(C)3084

(D)3088

(E)3094

25.How many 3-digit numbers do not have an even digit or a zero?

(A)20

(B)30

(C)60

(D)80

(E)125

Hard

27.Which one of the following is the minimum value of the sum of two integers whose product is 36?

(A)37

(B)20

(C)15

(D)13

(E)12

28. If a and b are positive integers, and x = 2 3 7 a, and y = 2 2 8 b, and the values of both x and y lie between 120 and 130 (not including the two), then a – b =

(A)–2

(B)–1

(C)0

(D)1

(E)2

31.a, b, c, d, and e are five consecutive numbers in increasing order of size. Deleting one of the five numbers from the set decreased the sum of the remaining numbers in the set by 20%. Which one of the following numbers was deleted?

(A)a

(B)b

(C)c

(D)d

(E)e

32.A set has exactly five consecutive positive integers starting with 1. What is the percentage decrease in the average of the numbers when the greatest one of the numbers is removed from the set?

(A)5

(B)8.5

(C)12.5

(D)15.2

(E)16.66

46GRE Math Bible

Answers and Solutions to Problem Set E

Easy

1.We are given that x is divisible by 3 and 4. Hence, x must be a common multiple of 3 and 4. The least common multiple of 3 and 4 is 12. So, x is a multiple of 12. The answer is (B).

2.Numbers ending with 0, 1, 5, or 6 will have their squares also ending with the same digit.

For example,

10 ends with 0, and 102 (= 100) also ends with 0. 11 ends with 1, and 112 (= 121) also ends with 1. 15 ends with 5, and 152 (= 225) also ends with 5. 16 ends with 6, and 162 (= 256) also ends with 6.

Among the four numbers 0, 1, 5, or 6, even numbers only end with 0 or 6. Choice (D) has one such number. The answer is (D).

3. A number divisible by 2 ends with one of the digits 0, 2, 4, 6, or 8.

If a number is divisible by 3, then the sum of its digits is also divisible by 3.

Hence, a number divisible by both 2 and 3 will follow both of the above rules.

Choices (A) and (C) do not end with an even digit. Hence, eliminate them.

The sum of digits of Choice (B) is 1 + 2 + 9 + 6 = 18, which is divisible by 3. Also, the last digit is 6. Hence, choice (B) is correct.

Next, the sum of the digits of choices (D) and (E) are 1 + 4 + 0 + 6 (= 11) and 1 + 4 + 1 + 4 (= 10), respectively, and neither is divisible by 3. Hence, reject the two choices.

Hence, the answer is (B).

4.A prime number is divisible by no other numbers, but itself and 1. Hence, the only prime number divisible by 2 is 2 itself; and the only prime number divisible by 3 is 3 itself. Since the number of primes in each column is 1, the answer is (C).

5.Choice (A): 11 6 can be factored as 11 2 3. The product of more than two primes. Reject. Choice (B): 13 22 can be factored as 13 2 11. The product of more than two primes. Reject. Choice (C): 14 23 can be factored as 7 2 23. The product of more than two primes. Reject. Choice (D): 17 21 can be factored as 17 3 7. The product of more than two primes. Reject. Choice (E): 13 23 cannot be further factored and is itself the product of two primes. Accept.

The answer is (E).

6.The smallest prime number greater than 21 is 23, and the largest prime number less than 16 is 13. The product of the two is 13 23, which is listed in choice (C). The answer is (C).

7.Suppose b equals 0. Then a = b + 3 = 0 + 3 = 3. Hence, Column A equals ab = 30. Here, Column A is less than Column B, which equals 40.

Now, suppose b equals 1. Then a = b + 3 = 1 + 3 = 4. Hence, Column A equals ab = 41. Here Column A is greater than Column B, which equals 40.

Hence, we have a double case, and the answer is (D).

8. The last digit of the number 252 (in Column A) is 2, and the last digit of the number 152 (in Column B) is also 2. Hence, both numbers raised to the same power (here 56) should end with the same digit. So, 15256 should end with the same digit as 25256. The answer is (C). A small example: the last digit of 62 (= 36) is 6 and the last digit of 162 (= 256) is also 6.

Medium

9.We have that p/q = 7.15. Solving for p yields p = 7.15q = 7q + 0.15q = 7(a positive integer) + 0.15q = (a positive integer) + 0.15q. Now, p is a positive integer only when 0.15q is an integer. Now, 0.15q equals 15/100 q = 3q/20 and would result in an integer only when the denominator of the fraction (i.e., 20) is canceled out by q. This happens only when q is a multiple of 20. Hence, q = 20, or 40, or 60, …. Any of these values is greater than 15. Hence, Column A is greater than Column B, and the answer is (A).

10.Column A: Since all the digits of the 3-digit number X are greater than 4, each digit must be greater than or equal to 5. Hence, the sum of the three digits must be greater than or equal to 3 × 5 = 15. Also, since

the maximum value of each digit is 9, the maximum possible value of the sum of its digits is 3 × 9 = 27. Hence, we have the inequality 15 ≤ Column A ≤ 27.

Column B: Since all the digits in the 4-digit number Y are less than 5, each digit must be less than or equal to 4. Hence, the sum of the four digits must be less than or equal to 4 × 4 = 16. Also, the minimum value of the sum of the digits of a 4-digit number is 1 (for example, for 1000, the sum of the digits is 1). Hence, we have the inequality 1 ≤ Column B ≤ 16.

Since the inequality for Column A, 15 ≤ Column A ≤ 27, and the inequality for Column B, 1 ≤ Column B ≤ 16, have a common range (numbers between 15 and 16, inclusive, satisfy both inequalities), an inequality between the two columns cannot be derived. Hence, we cannot know which column is greater. The answer is (D).

11.Any number that is divisible by the five numbers 2, 3, 4, 5, and 6 must also be a multiple of all five numbers. So, Column A (refers to the set of numbers that are divisible by 2, 3, 4, 5, and 6) and Column B (refers to the set of numbers that are a multiple of 2, 3, 4, 5 and 6) refer to the same set. Hence, the least values of the two columns are the same. So, Column A equals Column B, and the answer is (C).

12.Since 3 divides b with a result of 2 and a remainder of 1, b = 2 3 + 1 = 7. Since number 3 divides a with a result of b (which we now know equals 7) and a remainder of 2, a = b 3 + 2 = 7 3 + 2 = 23. The answer is (D).

13.Choice (A): 45/7 = 6 + 3/7, so x = 3. Now, 45/14 = 3 + 3/14. The remainder is 3, not x + 7 (= 10). Reject.

Choice (B): 53/7 = 7 + 4/7, so x = 4. Now, 53/14 = 3 + 11/14. The remainder is 11, and equals x + 7 (= 11). Accept the choice.

Choice (C): 72/7 = 10 + 2/7, so x = 2. Now, 72/14 = 5 + 2/14. The remainder is 2, not x + 7 (= 9). Reject. Choice (D): 85/7 = 12 + 1/7, so x = 1. Now, 85/14 = 6 + 1/14. The remainder is 1, not x + 7 (= 8). Reject. Choice (E): 100/7 = 14 + 2/7, so x = 2. Now, 100/14 = 7 + 2/14. The remainder is 2, not x + 7 (= 9). Reject.

The answer is (B).

14.Choice (A) = 5.43 + 4.63 – 3.24 – 2.32 = 4.5.

Choice (B) = 5.53 + 4.73 – 3.34 – 2.42 = 4.5 = Choice (A). Reject choices (A) and (B). Choice (C) = 5.53 + 4.53 – 3.34 – 2.22 = 4.5 = Choice (A). Reject choice (C).

Choice (D) = 5.43 + 4.73 – 3.24 – 2.42 = 4.5 = Choice (A). Reject choice (D).

Choice (E) = 5.43 + 4.73 – 3.14 – 2.22 = 4.8. Correct.

The answer is (E).

48GRE Math Bible

15.Since each of the two integers a and b ends with the same digit, the difference of the two numbers ends with 0. For example 642 – 182 = 460, and 460 ends with 0. Hence, the answer is (A).

16.Suppose a and b are the two digits of the number x, and let a represent the greater of the two. From the given information, we have

Applying the Difference of Squares formula, a2 – b2 = (a – b)(a + b), to the second equation yields

Adding equations (1) and (2) yields 2a = 16. Dividing by 2 yields a = 8. Substituting this in equation (2) yields 8 + b = 10. Solving for b yields b = 2. Hence, the two digits are 2 and 8. They can be arranged in any order. Hence, 28 and 82 are two feasible solutions. Now, 28 is less than Column B, and 82 is greater than Column B. Hence, we have a double case, and the answer is (D).

17.Since each one of the three numbers a, b, and c is divisible by 3, the numbers can be represented as 3p, 3q, and 3r, respectively, where p, q, and r are integers. The product of the three numbers is 3p 3q 3r = 27(pqr). Since p, q, and r are integers, pqr is an integer and therefore abc is divisible by 27. The answer is

(B).

18.Let the digits of the five-digit positive number be represented by 5 compartments:

Each of the last four compartments can be filled in 3 ways (with any one of the numbers 0, 3 and 5).

The first compartment can be filled in only 2 ways (with only 3 and 5, not with 0, because placing 0 in the first compartment would yield a number with fewer than 5 digits).

Hence, the total number of ways the five compartments can be filled in is 2 3 3 3 3 = 162. The answer is (D).

19. Let m be a number that has a remainder of 3 when divided by any of the numbers 4, 5, and 6. Then m – 3 must be exactly divisible by all three numbers. Hence, m – 3 must be a multiple of the Least Common Multiple of the numbers 4, 5, and 6. The LCM is 3 4 5 = 60. Hence, we can suppose m – 3 = 60p, where p is a positive integer. Replacing p with n, we get m – 3 = 60n. So, m = 60n + 3. Choice (E) is in the same format 120n + 3 = 60(2n) + 3. Hence, the answer is (E).

We can also subtract 3 from each answer-choice, and the correct answer will be divisible by 60:

Choice (A): If n = 1, then (12n + 3) – 3 = 12n = 12, not divisible by 60. Reject.

Choice (B): If n = 1, then (24n + 3) – 3 = 24n = 24, not divisible by 60. Reject.

Choice (C): If n = 1, then (80n + 3) – 3 = 80n = 80, not divisible by 60. Reject.

Choice (D): If n = 1, then (90n + 2) – 3 = 90n – 1 = 89, not divisible by 60. Reject.

Choice (E): (120n + 3) – 3 = 120n, divisible by 60 for any integer n. Hence, correct.

Number Theory 49

20. Suppose the two positive integers m and n do not have a common factor, apart from 1. Then the LCM of m and n is mn. For example, when m = 14 (= 2 • 7) and n = 15 (= 3 • 5), the LCM is 14 • 15. In this case, the LCM equals mn, and the Column A equals Column B.

Now, suppose the two integers m and n have at least one common factor (other than 1). Then the LCM uses the common factors only once, unlike mn. Hence, the LCM is less than mn. For example, suppose m = 10 (= 2 • 5) and n = 14 (= 2 • 7). Here, m and n have 2 as a common factor. The LCM of m and n is 2 • 5 • 7 = 70, and mn equals 2 • 5 • 2 • 7 = 140. The LCM did not use the underlined 2 in the evaluation. Hence, here, Column A is less than Column B.

Since this is a double case, the answer is (D).

21. Any number divisible by both 6 and 8 must be a multiple of the least common multiple of the two numbers, which is 24. Hence, any such number can be represented as 24n. If 3072 is one such number and is represented as 24n, then the next such number should be 24(n + 1) = 24n + 24 = 3072 + 24 = 3096. The answer is (E).

22.Any number divisible by both m and n must be a multiple of the least common multiple of the two numbers, which is given to be 24. The first multiple of 24 greater than 3070 is 3072. Hence, the answer is

(A).

23.The least common multiple of 6 and 8 is 24. Hence, a multiple of both 6 and 8 must also be a multiple of 24.

We know that a multiple of 24 exists once in every 24 consecutive numbers. For example, between the numbers 34 (not itself a multiple of 24) and 34 + 24, there must be exactly one multiple of 24. Similarly, exactly one multiple of 24 must exist between 300 (not itself a multiple of 24) and 300 + 24 (= 324, Column B), and this number is the first multiple of 24 larger than 300, Column A. So, Column A lies between 300 and 324 (Column B) and therefore is less than Column B. The answer is (B).

24. The consecutive integers a, b, and c in the increasing order of size can be expressed as a, a + 1, a + 2, respectively.

Column A + 1

30

Since Column B is 1/30 units greater than Column A, the answer is (B).

50GRE Math Bible

25.There are 5 digits that are not even or zero: 1, 3, 5, 7, and 9. Now, let’s count all the three-digit numbers that can be formed from these five digits. The first digit of the number can be filled in 5 ways with

any one of the mentioned digits. Similarly, the second and third digits of the number can be filled in 5 ways. Hence, the total number of ways of forming the three-digit number is 125 (= 5 × 5 × 5). The answer is (E).

26.Since the digits of the number differ by 4, the number can be any one of the numbers 15, 26, 37, 48, and 59 or any one of their reverse numbers. Hence, the positive difference between the digits varies from

52– 12 = 24 (for the numbers 15 or 51) to 92 – 52 = 56 (for the numbers 59 or 95). Hence, Column A is a number between 24 and 56, inclusive, and therefore is always greater than 15, the value of Column B. Hence, the answer is (A).

Hard

27. List all possible factors x and y whose product is 36, and calculate the corresponding sum x + y:

From the table, the minimum sum is 12. The answer is (E).

28. We are given that x = 2 3 7 a = 42a and y = 2 2 8 b = 32b.

We are given that the values of both x and y lie between 120 and 130 (not including the two).

The only multiple of 42 in this range is 42 × 3 = 126. Hence, x = 126 and a = 3. The only multiple of 32 in this range is 32 × 4 = 128. Hence, y = 128 and b = 4. Hence, a – b = 3 – 4 = –1. The answer is (B).

29. Usually, a system having a single constraint such as 5m + 7n = 46 will not have a unique solution.

But the given system has two constraints:

1)5m + 7n = 46

2)m and n are positive integers

Hence, we might have a unique solution. Let’s see:

Let p = 5m (p is a multiple of 5) and q = 7n (q is a multiple of 7) such that and p + q = 46. Subtracting q from both sides yields p = 46 – q [(a positive multiple of 5) equals 46 – (a positive multiple of 7)]. Let’s look at how many such solutions exist:

If q = 7, p = 46 – 7 = 39, not a multiple of 5. Reject.

If q = 14, p = 46 – 14 = 32, not a multiple of 5. Reject.

If q = 21, p = 46 – 21 = 25, a multiple of 5. Acceptable. So, m = 25/5 = 5 and q = 21/7 = 3.

If q = 28, p = 46 – 28 = 18, not a multiple of 5. Reject.

If q = 35, p = 46 – 35 = 11, not a multiple of 5. Reject.

If q = 42, p = 46 – 42 = 4, not a multiple of 5. Reject.

If q ≥ 49, p ≤ 46 – 49 = –3, not positive. Reject.

The solution is Column A = m = 5, and Column B = n = 3. Hence, Column A is greater and the answer is

(A).