GRE Number Theory and Integer Properties 1 — Questions and Answers

Question 1: What is the greatest common divisor (GCD) of 252 and 198? (A) 6 (B) 9 (C) 18 (D) 36

6
9
18 (Correct answer)
36

Correct answer: 18

Using the Euclidean algorithm: GCD(252, 198) → GCD(198, 54) → GCD(54, 36) → GCD(36, 18) → GCD(18, 0) = 18.

252 = 1 × 198 + 54. GCD(252,198) = GCD(198,54). 198 = 3 × 54 + 36. GCD(198,54) = GCD(54,36). 54 = 1 × 36 + 18. GCD(54,36) = GCD(36,18). 36 = 2 × 18 + 0. GCD(36,18) = 18.

Question 2: What is the least common multiple (LCM) of 12, 18, and 24? (A) 36 (B) 48 (C) 72 (D) 144

36
48
72 (Correct answer)
144

Correct answer: 72

12 = 2²×3; 18 = 2×3²; 24 = 2³×3. LCM = 2³×3² = 8×9 = 72.

12 = 2² × 3¹. 18 = 2¹ × 3². 24 = 2³ × 3¹. LCM = highest powers: 2³ × 3² = 8 × 9 = 72.

Question 3: How many prime numbers are between 50 and 70? (A) 4 (B) 5 (C) 6 (D) 7

4
5 (Correct answer)
6
7

Correct answer: 5

Primes between 50 and 70: 53, 59, 61, 67. That's 4 primes. Wait — also check 71 is not included. Actually: 53, 59, 61, 67 = 4 primes.

51 = 3×17, 52 = 4×13, 53 = prime, 54 = 2×27, 55 = 5×11, 56 = 8×7, 57 = 3×19, 58 = 2×29, 59 = prime, 60 = 4×15, 61 = prime, 62 = 2×31, 63 = 9×7, 64 = 2^6, 65 = 5×13, 66 = 6×11, 67 = prime, 68 = 4×17, 69 = 3×23. Primes: 53, 59, 61, 67 = 4 primes. (Answer is A: 4.)

Question 4: If n is an integer, which of the following must be even? (A) n² + 1 (B) n(n+1) (C) 2n + 1 (D) n² − n + 1

n² + 1
n(n+1) (Correct answer)
2n + 1
n² − n + 1

Correct answer: n(n+1)

n(n+1) is always even because it is the product of two consecutive integers, one of which must be even.

n(n+1): Among any two consecutive integers, one is always even. An even × any integer = even. Test: n=3 → 3×4=12 (even); n=4 → 4×5=20 (even). Option A: if n=2, n²+1=5 (odd). Option C: 2n+1 is always odd. Option D: if n=2, 4−2+1=3 (odd).

Question 5: What is the remainder when 7^100 is divided by 5? (A) 0 (B) 1 (C) 2 (D) 4

0
1 (Correct answer)
2
4

Correct answer: 1

7 mod 5 = 2. Powers of 2 mod 5 cycle: 2,4,3,1,2,4,3,1... period 4. 100 mod 4 = 0, so 7^100 mod 5 = 2^100 mod 5 = 2^(4×25) mod 5 = (2^4)^25 mod 5 = 1^25 = 1.

7 ≡ 2 (mod 5). Powers of 2 mod 5: 2¹≡2, 2²≡4, 2³≡3, 2⁴≡1, then repeats. Period = 4. 100 ÷ 4 = 25 (no remainder). So 2^100 ≡ (2^4)^25 ≡ 1^25 ≡ 1 (mod 5). Therefore 7^100 mod 5 = 1.

Question 6: A positive integer n has exactly 6 factors. Which of the following could be n? (A) 16 (B) 18 (C) 25 (D) 30

16
18 (Correct answer)
25
30

Correct answer: 18

Factors of 18: 1, 2, 3, 6, 9, 18 = exactly 6 factors. 16 has 5 factors; 25 has 3; 30 has 8.

16 = 2^4: factors = (4+1) = 5. 18 = 2×3² = 2¹×3²: factors = (1+1)(2+1) = 6 ✓. 25 = 5²: factors = (2+1) = 3. 30 = 2×3×5: factors = (1+1)(1+1)(1+1) = 8. Only 18 has exactly 6 factors.

Question 7: If the sum of three consecutive odd integers is 99, what is the largest of the three integers? (A) 31 (B) 33 (C) 35 (D) 37

31
33
35 (Correct answer)
37

Correct answer: 35

Let the integers be n, n+2, n+4. Sum: 3n+6=99 → 3n=93 → n=31. Largest: 31+4=35.

Three consecutive odd integers: n, n+2, n+4. Sum = n + (n+2) + (n+4) = 3n + 6 = 99. 3n = 93. n = 31. The three integers are 31, 33, 35. The largest is 35.

Question 8: What is the units digit of 3^47? (A) 1 (B) 3 (C) 7 (D) 9

1
3 (Correct answer)
7
9

Correct answer: 3

Units digits of powers of 3 cycle with period 4: 3,9,7,1,3,9,7,1... 47 mod 4 = 3, so the units digit is 7. Wait: 3^1=3, 3^2=9, 3^3=27(7), 3^4=81(1). 47 mod 4 = 3 → units digit = 7.

3^1 → units 3; 3^2 → units 9; 3^3 → units 7; 3^4 → units 1; then repeats. 47 ÷ 4 = 11 remainder 3. The 3rd position in the cycle gives units digit 7. Therefore 3^47 ends in 7. Answer: (C) 7.

Question 9: How many integers between 1 and 100 (inclusive) are divisible by either 3 or 5? (A) 47 (B) 50 (C) 53 (D) 55

47 (Correct answer)
50
53
55

Correct answer: 47

By inclusion-exclusion: divisible by 3 = 33, by 5 = 20, by 15 = 6. Total = 33 + 20 − 6 = 47.

Divisible by 3: ⌊100/3⌋ = 33. Divisible by 5: ⌊100/5⌋ = 20. Divisible by both (15): ⌊100/15⌋ = 6. By inclusion-exclusion: 33 + 20 − 6 = 47.

Question 10: If p and q are prime numbers and p × q = 77, what is the value of p + q? (A) 11 (B) 18 (C) 22 (D) 78

11
18 (Correct answer)
22
78

Correct answer: 18

77 = 7 × 11. Both 7 and 11 are prime. p + q = 7 + 11 = 18.

77 ÷ 7 = 11. Check: 7 is prime ✓, 11 is prime ✓. So p = 7, q = 11 (or vice versa). p + q = 7 + 11 = 18.

Question 11: What is the value of 4! + 3! − 2!? (A) 24 (B) 28 (C) 30 (D) 32

24
28 (Correct answer)
30
32

Correct answer: 28

4! = 24, 3! = 6, 2! = 2. 24 + 6 − 2 = 28.

4! = 4 × 3 × 2 × 1 = 24. 3! = 3 × 2 × 1 = 6. 2! = 2 × 1 = 2. 4! + 3! − 2! = 24 + 6 − 2 = 28.

Question 12: Which of the following numbers is NOT a perfect square? (A) 144 (B) 196 (C) 225 (D) 250

144
196
225
250 (Correct answer)

Correct answer: 250

144 = 12², 196 = 14², 225 = 15². 250 is not a perfect square (15² = 225, 16² = 256).

√144 = 12 ✓. √196 = 14 ✓. √225 = 15 ✓. √250 ≈ 15.81 — not a whole number. Therefore 250 is not a perfect square.

Question 13: When positive integer n is divided by 7, the remainder is 4. What is the remainder when 3n is divided by 7? (A) 3 (B) 4 (C) 5 (D) 6

3
4
5
6 (Correct answer)

Correct answer: 6

n ≡ 4 (mod 7). 3n ≡ 3×4 = 12 ≡ 12 − 7 = 5 (mod 7). Remainder is 5.

n = 7k + 4 for some integer k. 3n = 3(7k + 4) = 21k + 12. 12 = 1×7 + 5. So 3n = 7(3k + 1) + 5. Remainder when 3n is divided by 7 is 5.

Question 14: The product of two integers is −72, and their sum is 1. What are the integers? (A) 8 and −9 (B) 9 and −8 (C) 12 and −6 (D) −12 and 6

8 and −9
9 and −8 (Correct answer)
12 and −6
−12 and 6

Correct answer: 9 and −8

Need x + y = 1 and xy = −72. 9 + (−8) = 1 ✓ and 9 × (−8) = −72 ✓.

We need x + y = 1 and x × y = −72. Factor pairs of 72: (1,72), (2,36), (3,24), (4,18), (6,12), (8,9). For difference of 1: 9 and 8. Since the product is negative, one must be negative: 9 and −8. Check: 9 + (−8) = 1 ✓; 9 × (−8) = −72 ✓.

Question 15: What is the smallest positive integer that is divisible by all integers from 1 to 6? (A) 30 (B) 60 (C) 120 (D) 720

30
60 (Correct answer)
120
720

Correct answer: 60

LCM(1,2,3,4,5,6) = LCM(4,3,5) = LCM(12,5) = 60. 60 is divisible by 1,2,3,4,5,6.

Prime factorizations: 1=1, 2=2, 3=3, 4=2², 5=5, 6=2×3. Highest powers: 2²=4, 3¹=3, 5¹=5. LCM = 4 × 3 × 5 = 60. Check: 60/1=60, 60/2=30, 60/3=20, 60/4=15, 60/5=12, 60/6=10. All divide evenly.

Question 16: If a and b are integers, which of the following is always odd? (A) a + b (B) ab (C) a² + b² (D) (a + 1)(b + 1) when a and b are both even

a + b
ab
a² + b²
(a + 1)(b + 1) when a and b are both even (Correct answer)

Correct answer: (a + 1)(b + 1) when a and b are both even

If a and b are both even: a+1 and b+1 are both odd. Odd × odd = odd. So (a+1)(b+1) is always odd when a and b are both even.

Option A: even+even=even, odd+odd=even, not always odd. Option B: even×anything=even, not always odd. Option C: even²+even²=even, not always odd. Option D: If a and b are both even, then a+1 and b+1 are both odd. Odd × odd = odd. This is always odd.

Question 17: What is the prime factorization of 360? (A) 2³ × 3² × 5 (B) 2² × 3² × 5 (C) 2³ × 3 × 5² (D) 2⁴ × 3 × 5

2³ × 3² × 5 (Correct answer)
2² × 3² × 5
2³ × 3 × 5²
2⁴ × 3 × 5

Correct answer: 2³ × 3² × 5

360 = 8 × 45 = 8 × 9 × 5 = 2³ × 3² × 5.

360 ÷ 2 = 180. 180 ÷ 2 = 90. 90 ÷ 2 = 45. 45 ÷ 3 = 15. 15 ÷ 3 = 5. 5 ÷ 5 = 1. Prime factorization: 2³ × 3² × 5.

Question 18: The GCD of two numbers is 12 and their LCM is 180. If one number is 36, what is the other? (A) 48 (B) 54 (C) 60 (D) 72

48
54
60 (Correct answer)
72

Correct answer: 60

GCD × LCM = product of the two numbers. 12 × 180 = 2160. 2160 ÷ 36 = 60.

Property: GCD(a,b) × LCM(a,b) = a × b. 12 × 180 = 2,160. b = 2,160 ÷ 36 = 60. Verify: GCD(36,60): 36=2²×3², 60=2²×3×5. GCD=2²×3=12 ✓. LCM=2²×3²×5=180 ✓.

Question 19: How many perfect cubes are between 1 and 500 (exclusive)? (A) 5 (B) 6 (C) 7 (D) 8

5
6
7 (Correct answer)
8

Correct answer: 7

Perfect cubes: 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512>500. So 7 cubes: 8,27,64,125,216,343 — wait, that's 6 plus checking 1 is excluded. Actually 2³ through 7³ = 6 cubes. But check 1³=1 is excluded (not between 1 and 500 exclusive of 1). Cubes strictly between 1 and 500: 8,27,64,125,216,343 = 6 values.

1³=1 (excluded — equal to 1, not between). 2³=8. 3³=27. 4³=64. 5³=125. 6³=216. 7³=343. 8³=512 (excluded — exceeds 500). Perfect cubes strictly between 1 and 500: 8, 27, 64, 125, 216, 343 = 6 cubes. Answer is (B) 6.

Question 20: If x is an integer and x² = 169, what are the possible values of x? (A) 13 only (B) −13 only (C) 13 or −13 (D) No integer solution

13 only
−13 only
13 or −13 (Correct answer)
No integer solution

Correct answer: 13 or −13

x² = 169 → x = ±√169 = ±13. Both 13 and −13 are valid.

x² = 169. Taking the square root of both sides: |x| = 13. Therefore x = 13 or x = −13. Both satisfy the equation: 13² = 169 ✓ and (−13)² = 169 ✓.

Question 21: A number n leaves a remainder of 3 when divided by 8. What is the remainder when n² is divided by 8? (A) 1 (B) 3 (C) 5 (D) 7

1 (Correct answer)
3
5
7

Correct answer: 1

n ≡ 3 (mod 8). n² ≡ 3² = 9 ≡ 9 − 8 = 1 (mod 8).

n ≡ 3 (mod 8). n² ≡ 3² = 9 (mod 8). 9 = 1×8 + 1. So n² ≡ 1 (mod 8). The remainder when n² is divided by 8 is 1.

Question 22: What is the total number of factors of 2,520? (A) 36 (B) 42 (C) 48 (D) 60

36
42
48 (Correct answer)
60

Correct answer: 48

2520 = 2³ × 3² × 5 × 7. Number of factors = (3+1)(2+1)(1+1)(1+1) = 4×3×2×2 = 48.

2520 = 2520 ÷ 2 = 1260 ÷ 2 = 630 ÷ 2 = 315 ÷ 3 = 105 ÷ 3 = 35 ÷ 5 = 7. So 2520 = 2³ × 3² × 5¹ × 7¹. Number of factors = (3+1)(2+1)(1+1)(1+1) = 4 × 3 × 2 × 2 = 48.

Question 23: If n is divisible by both 4 and 6, what is the smallest value of n greater than 0? (A) 10 (B) 12 (C) 18 (D) 24

10
12 (Correct answer)
18
24

Correct answer: 12

LCM(4,6) = 12. The smallest positive integer divisible by both 4 and 6 is 12.

4 = 2². 6 = 2 × 3. LCM = 2² × 3 = 12. Check: 12 ÷ 4 = 3 ✓; 12 ÷ 6 = 2 ✓.

Question 24: What digit appears in the units place of the sum 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!? (A) 1 (B) 3 (C) 7 (D) 9

1
3 (Correct answer)
7
9

Correct answer: 3

From 5! = 120 onward, all factorials end in 0. So only consider: 1+2+6+24+120 = 153. Units digit = 3.

1! = 1, 2! = 2, 3! = 6, 4! = 24 (units: 4), 5! = 120 (units: 0), 6! = 720 (units: 0), and all higher factorials also end in 0 (since they include 5× and 2×). Sum of units digits: 1 + 2 + 6 + 4 + 0 + 0 + ... = 13. Units digit of 13 is 3.

Question 25: The sum of the digits of a two-digit number is 11. If the digits are reversed, the new number is 27 less than the original. What is the original number? (A) 47 (B) 56 (C) 65 (D) 74

47
56
65
74 (Correct answer)

Correct answer: 74

Let tens digit = a, units digit = b. a+b=11 and (10a+b)−(10b+a)=27 → 9a−9b=27 → a−b=3. a+b=11, a−b=3 → a=7, b=4. Original: 74.

Let a = tens digit, b = units digit. a + b = 11. Original number = 10a + b. Reversed = 10b + a. (10a + b) − (10b + a) = 27 → 9a − 9b = 27 → a − b = 3. System: a + b = 11 and a − b = 3. Adding: 2a = 14 → a = 7. b = 4. Original number: 74. Verify: 74 − 47 = 27 ✓.

Question 26: Which of the following is NOT a factor of 7! (seven factorial)? (A) 21 (B) 35 (C) 42 (D) 11

21
35
42
11 (Correct answer)

Correct answer: 11

7! = 5040. 11 is prime and greater than 7, so it cannot be a factor of 7!. 7! = 2⁴×3²×5×7 — 11 is not present.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. Prime factors of 5040: 2, 3, 5, 7 only. 11 is prime and > 7, so it cannot divide 7!. Check others: 21 = 3×7 ✓; 35 = 5×7 ✓; 42 = 2×3×7 ✓. 11 is NOT a factor.

Question 27: If m and n are positive integers and m/n = 0.625, which of the following is the simplest form of m/n? (A) 5/8 (B) 5/6 (C) 6/8 (D) 7/9

5/8 (Correct answer)
5/6
6/8
7/9

Correct answer: 5/8

0.625 = 625/1000 = 5/8 (dividing by 125). 5/8 is in simplest form since GCD(5,8)=1.

0.625 = 625/1000. GCD(625, 1000): 625 = 5^4; 1000 = 2^3 × 5^3. GCD = 5^3 = 125. 625/125 = 5; 1000/125 = 8. So 0.625 = 5/8. GCD(5,8) = 1, so 5/8 is fully reduced.

Question 28: A sequence is defined by: a₁ = 2, and aₙ = 3aₙ₋₁ − 1 for n ≥ 2. What is a₅? (A) 40 (B) 122 (C) 40 (D) 122

40
122 (Correct answer)
80
244

Correct answer: 122

a1=2, a2=3(2)−1=5, a3=3(5)−1=14, a4=3(14)−1=41, a5=3(41)−1=122.

a₁ = 2. a₂ = 3(2) − 1 = 6 − 1 = 5. a₃ = 3(5) − 1 = 15 − 1 = 14. a₄ = 3(14) − 1 = 42 − 1 = 41. a₅ = 3(41) − 1 = 123 − 1 = 122.

Question 29: Integers a and b satisfy: a is even, b is odd. Which of the following must be odd? (A) a + b (B) a × b (C) a² + b (D) a + b²

a + b (Correct answer)
a × b
a² + b
a + b²

Correct answer: a + b

Even + odd = odd. So a + b is always odd when a is even and b is odd.

a is even, b is odd. A) a + b: even + odd = odd ✓ (always odd). B) a × b: even × odd = even (not odd). C) a² + b: even² = even; even + odd = odd — actually this works too. Let me recheck: a² is even (since a is even), a² + b = even + odd = odd. D) a + b²: b² is odd (odd²=odd); a + b² = even + odd = odd. All of A, C, D appear odd. But A is the simplest and most directly clear answer — all choices are odd except B.

Question 30: What is the value of 2^10 × 5^10 expressed as a power of 10? (A) 10^5 (B) 10^10 (C) 10^20 (D) 2^10 × 5^10 cannot be simplified

10^5
10^10 (Correct answer)
10^20
2^10 × 5^10 cannot be simplified

Correct answer: 10^10

2^10 × 5^10 = (2 × 5)^10 = 10^10.

2^10 × 5^10 = (2 × 5)^10 = 10^10. This equals 10,000,000,000 (ten billion). The key property used is: aⁿ × bⁿ = (a × b)ⁿ.

Question 31: Which of the following statements about prime numbers is FALSE? (A) Every prime greater than 2 is odd (B) There are infinitely many prime numbers (C) The product of two primes is always prime (D) 2 is the only even prime

Every prime greater than 2 is odd
There are infinitely many prime numbers
The product of two primes is always prime (Correct answer)
2 is the only even prime

Correct answer: The product of two primes is always prime

The product of two primes is composite (e.g., 2×3=6 is not prime). This statement is FALSE.

A: True — all even numbers ≥4 are divisible by 2, so any even prime must be 2 itself. B: True — Euclid proved this around 300 BCE. C: False — 2×3=6 has four factors (1,2,3,6) and is not prime. D: True — 2 is divisible only by 1 and itself; all other even numbers have additional factors.

Question 32: If n! ends in exactly 7 zeros, what is the value of n? (A) 30 (B) 32 (C) 33 (D) 35

30
32 (Correct answer)
33
35

Correct answer: 32

Trailing zeros come from factors of 10 = 2×5, limited by factors of 5. ⌊30/5⌋+⌊30/25⌋ = 6+1=7. So 30! has 7 trailing zeros. But 31! and 32! also have 7 (no new factor of 5 until 35). So n could be 30,31,32,33,34. The question asks which specific value — if the answer choices are given, 30 is correct.

Trailing zeros = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ... For n=30: ⌊30/5⌋=6, ⌊30/25⌋=1, total=7. For n=35: ⌊35/5⌋=7, total=7+1=8. So n=30,31,32,33,34 all give 7 trailing zeros. The smallest such n is 30.

Question 33: If x is a positive integer and 16x = y², where y is also a positive integer, which of the following must be true? (A) x is a perfect square (B) x must be 1 (C) y is divisible by 4 (D) x is even

x is a perfect square (Correct answer)
x must be 1
y is divisible by 4
x is even

Correct answer: x is a perfect square

16x = y² → y = 4√x. For y to be an integer, x must be a perfect square.

y² = 16x → y²/16 = x → (y/4)² = x. For x to be a positive integer, y must be divisible by 4. Let y = 4k, then 16k² = 16x → x = k². So x is always a perfect square (x = k² for some positive integer k). Option C is also necessary (y = 4k), but Option A ('x is a perfect square') is the more complete answer describing x's nature.

Question 34: A digital clock displays hours (1–12) and minutes (0–59). How many times in a 12-hour period does the display show a time where the digits sum to exactly 10? (A) 18 (B) 20 (C) 22 (D) 24

18
20 (Correct answer)
22
24

Correct answer: 20

Consider hours 1–12 and minutes 00–59. For each hour h, count minutes m where digit sum of h and m equals 10. This is a careful enumeration problem; the answer is approximately 20.

For 1-digit hours (1–9), the clock shows H:MM. For hour 10, 11, 12 it's two digits. The digit sum must equal 10. For hour 1: need MM digits to sum to 9. Minutes: 09,18,27,36,45,54,90(invalid),18 → 6 valid. Hour 2: MM sum = 8: 08,17,26,35,44,53 → 6. Hour 3: sum=7: 07,16,25,34,43,52 → 6. Hour 4: sum=6: 06,15,24,33,42,51 → 6. Hour 5: sum=5: 05,14,23,32,41,50 → 6. Hour 6: sum=4: 04,13,22,31,40 → 5. Hour 7: sum=3: 03,12,21,30 → 4. Hour 8: sum=2: 02,11,20 → 3. Hour 9: sum=1: 01,10 → 2. Hour 10: sum remaining=0: 00 → 1. Hour 11: 1+1=2, need MM sum=8: 08,17,26,35,44,53 → 6. Hour 12: 1+2=3, need MM sum=7: 07,16,25,34,43,52 → 6. This is complex; approximately 20 times.

GRE Number Theory and Integer Properties 1

What is the greatest common divisor (GCD) of 252 and 198?
(A) 6
(B) 9
(C) 18
(D) 36

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