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    Hard GRE Integer Questions Practice Questions

    July 8, 202611 min read15 views
    Hard GRE Integer Questions Practice Questions

    Concept Explanation

    GRE integer questions focus on the properties of whole numbers, including divisibility, prime factorization, remainders, and parity. An integer is any number from the set { . . . , βˆ’ 3 , βˆ’ 2 , βˆ’ 1 , 0 , 1 , 2 , 3 , . . . } \{..., -3, -2, -1, 0, 1, 2, 3, ...\} , meaning it has no fractional or decimal component. At a hard difficulty level, the GRE Prep curriculum requires you to combine multiple properties simultaneously, such as identifying the number of trailing zeros in a large factorial or determining the units digit of an exponential expression. Understanding the Fundamental Theorem of Arithmetic is essential, as it states that every integer greater than 1 is either a prime number or can be represented as a unique product of prime numbers. This concept often serves as the backbone for solving complex problems involving least common multiples (LCM) and greatest common divisors (GCD).

    Key topics encountered in Hard GRE Integer Questions include:

    • Divisibility Rules: Advanced applications of rules for 3, 4, 7, 8, 9, and 11.
    • Remainders: Using the remainder theorem and cyclical patterns (cyclicity) to find the remainder of large powers.
    • Factorials: Analyzing the prime factors within n ! n! , particularly the number of factors of 5 to find trailing zeros.
    • Consecutive Integers: Leveraging the property that the product of n n consecutive integers is always divisible by n ! n! .

    Solved Examples

    Reviewing these step-by-step solutions will help you navigate the logic required for high-difficulty integer problems.

    1. Example 1: Units Digit
      What is the units digit of 7 35 + 3 21 7^{35} + 3^{21} ?
      1. Find the cycle of units digits for 7: 7 1 = 7 , 7 2 = 9 , 7 3 = 3 , 7 4 = 1 7^1=7, 7^2=9, 7^3=3, 7^4=1 . The cycle is 4 terms long: { 7 , 9 , 3 , 1 } \{7, 9, 3, 1\} .
      2. Divide the exponent 35 by the cycle length 4: 35 Γ· 4 = 8 35 \div 4 = 8 remainder 3. The 3rd term in the cycle is 3.
      3. Find the cycle for 3: 3 1 = 3 , 3 2 = 9 , 3 3 = 7 , 3 4 = 1 3^1=3, 3^2=9, 3^3=7, 3^4=1 . The cycle is 4 terms long: { 3 , 9 , 7 , 1 } \{3, 9, 7, 1\} .
      4. Divide the exponent 21 by 4: 21 Γ· 4 = 5 21 \div 4 = 5 remainder 1. The 1st term in the cycle is 3.
      5. Add the units digits: 3 + 3 = 6 3 + 3 = 6 . The units digit is 6.
    2. Example 2: Trailing Zeros
      How many zeros are at the end of 100 ! 100! ?
      1. Trailing zeros are created by pairs of prime factors 2 and 5. In any factorial, factors of 5 are rarer than factors of 2.
      2. Count multiples of 5 up to 100: ⌊ 100 / 5 f l o o r = 20 \lfloor 100/5 floor = 20 .
      3. Count multiples of 5 2 5^2 (25) up to 100: ⌊ 100 / 25 f l o o r = 4 \lfloor 100/25 floor = 4 .
      4. Sum these counts: 20 + 4 = 24 20 + 4 = 24 . There are 24 trailing zeros.
    3. Example 3: Divisibility and Variables
      If n n is an integer and n 3 βˆ’ n n^3 - n is divided by 6, what is the remainder?
      1. Factor the expression: n 3 βˆ’ n = n ( n 2 βˆ’ 1 ) = n ( n βˆ’ 1 ) ( n + 1 ) n^3 - n = n(n^2 - 1) = n(n-1)(n+1) .
      2. Recognize that ( n βˆ’ 1 ) , n , ( n + 1 ) (n-1), n, (n+1) are three consecutive integers.
      3. Among any three consecutive integers, at least one must be a multiple of 2 and exactly one must be a multiple of 3.
      4. Since the product is divisible by both 2 and 3, it must be divisible by 6.
      5. The remainder when a multiple of 6 is divided by 6 is 0.

    Practice Questions

    Test your skills with these Hard GRE Integer Questions. Ensure you pay close attention to constraints like "positive integer" or "even/odd."

    1. If x x and y y are positive integers such that x 2 βˆ’ y 2 = 17 x^2 - y^2 = 17 , what is the value of x 2 + y 2 x^2 + y^2 ?
    2. What is the remainder when 3 100 3^{100} is divided by 10?
    3. If k k is the smallest positive integer such that 1260 k 1260k is a perfect cube, what is the value of k k ?

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    Practice GRE Questions
    1. The sum of 5 consecutive even integers is 120. What is the greatest of these integers?
    2. If p p is a prime number greater than 3, what is the remainder when p 2 p^2 is divided by 12?
    3. How many positive integers less than 100 have exactly three factors?
    4. If m m and n n are integers and 2 m Γ— 5 n = 0.000125 2^m \times 5^n = 0.000125 , what is the value of m + n m + n ?
    5. What is the greatest power of 3 that divides 54 ! 54! ?
    6. If n n is a positive integer, which of the following must be even? n 2 + n + 1 n^2 + n + 1 , n ( n + 1 ) n(n+1) , n 2 βˆ’ n + 5 n^2 - n + 5 , or 2 n + 1 2n + 1 ?
    7. A certain integer A A has exactly 8 factors. If two of its factors are 21 and 35, what is the value of A A ?

    Answers & Explanations

    1. Answer: 145. Factor the difference of squares: ( x βˆ’ y ) ( x + y ) = 17 (x-y)(x+y) = 17 . Since 17 is prime, the factors must be 1 and 17. Thus, x βˆ’ y = 1 x-y=1 and x + y = 17 x+y=17 . Solving the system gives 2 x = 18 β†’ x = 9 2x=18 \rightarrow x=9 and y = 8 y=8 . 9 2 + 8 2 = 81 + 64 = 145 9^2 + 8^2 = 81 + 64 = 145 .
    2. Answer: 1. Dividing a number by 10 yields its units digit as the remainder. The units digit of 3 n 3^n follows the cycle { 3 , 9 , 7 , 1 } \{3, 9, 7, 1\} . Since 100 Γ· 4 100 \div 4 has a remainder of 0, we take the 4th term in the cycle, which is 1.
    3. Answer: 7350. Prime factorize 1260: 2 2 Γ— 3 2 Γ— 5 1 Γ— 7 1 2^2 \times 3^2 \times 5^1 \times 7^1 . For a perfect cube, every exponent in the prime factorization must be a multiple of 3. We need 2 1 2^1 , 3 1 3^1 , 5 2 5^2 , and 7 2 7^2 to reach 2 3 , 3 3 , 5 3 , 7 3 2^3, 3^3, 5^3, 7^3 . So, k = 2 Γ— 3 Γ— 25 Γ— 49 = 7350 k = 2 \times 3 \times 25 \times 49 = 7350 .
    4. Answer: 28. Let the integers be x βˆ’ 4 , x βˆ’ 2 , x , x + 2 , x + 4 x-4, x-2, x, x+2, x+4 . Their sum is 5 x = 120 5x = 120 , so x = 24 x = 24 . The greatest is 24 + 4 = 28 24 + 4 = 28 .
    5. Answer: 1. Any prime p > 3 p > 3 can be written as 6 k Β± 1 6k \pm 1 . Squaring this gives ( 6 k Β± 1 ) 2 = 36 k 2 Β± 12 k + 1 (6k \pm 1)^2 = 36k^2 \pm 12k + 1 . Both 36 k 2 36k^2 and 12 k 12k are divisible by 12, leaving a remainder of 1.
    6. Answer: 4. An integer has exactly three factors if and only if it is the square of a prime number. Prime squares less than 100 are 2 2 = 4 , 3 2 = 9 , 5 2 = 25 , 7 2 = 49 2^2=4, 3^2=9, 5^2=25, 7^2=49 . (Note: 1 1 2 = 121 11^2=121 is too large).
    7. Answer: -9. Write 0.000125 as a fraction: 125 1 , 000 , 000 = 1 8 , 000 \frac{125}{1,000,000} = \frac{1}{8,000} . 8 , 000 = 8 Γ— 1000 = 2 3 Γ— ( 2 3 Γ— 5 3 ) = 2 6 Γ— 5 3 8,000 = 8 \times 1000 = 2^3 \times (2^3 \times 5^3) = 2^6 \times 5^3 . Thus, 2 m Γ— 5 n = 2 βˆ’ 6 Γ— 5 βˆ’ 3 2^m \times 5^n = 2^{-6} \times 5^{-3} . m = βˆ’ 6 , n = βˆ’ 3 m = -6, n = -3 . m + n = βˆ’ 9 m+n = -9 .
    8. Answer: 26. Using Legendre's Formula, count multiples of 3, 9, 27, etc.: ⌊ 54 / 3 f l o o r + ⌊ 54 / 9 f l o o r + ⌊ 54 / 27 f l o o r = 18 + 6 + 2 = 26 \lfloor 54/3 floor + \lfloor 54/9 floor + \lfloor 54/27 floor = 18 + 6 + 2 = 26 .
    9. Answer: n ( n + 1 ) n(n+1) . The product of two consecutive integers is always even because one of them must be even. For more practice on parity, try our GRE Practice Questions with Explanations.
    10. Answer: 105. The factors 21 and 35 imply A A is divisible by 3 , 5 , 3, 5, and 7 7 . The prime factors are 3 Γ— 5 Γ— 7 3 \times 5 \times 7 . The number of factors for p a Γ— q b Γ— r c p^a \times q^b \times r^c is ( a + 1 ) ( b + 1 ) ( c + 1 ) (a+1)(b+1)(c+1) . For 3 1 Γ— 5 1 Γ— 7 1 3^1 \times 5^1 \times 7^1 , the number of factors is 2 Γ— 2 Γ— 2 = 8 2 \times 2 \times 2 = 8 . Thus, A = 105 A = 105 .
    Interactive quizQuestion 1 of 5

    1. If \( x \) is an even integer and \( y \) is an odd integer, which of the following must be an odd integer?

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    Frequently Asked Questions

    What makes an integer question "hard" on the GRE?

    Hard integer questions typically involve large exponents, multiple variables with specific constraints, or properties like the number of factors and trailing zeros. They often require a combination of logic and number theory rather than simple arithmetic.

    How do I find the number of factors of a large integer?

    First, find the prime factorization of the number p 1 a Γ— p 2 b Γ— p 3 c . . . p_1^a \times p_2^b \times p_3^c... . The total number of factors is found by adding 1 to each exponent and multiplying them together: ( a + 1 ) ( b + 1 ) ( c + 1 ) . . . (a+1)(b+1)(c+1)... .

    Is 1 considered a prime number on the GRE?

    No, by definition, 1 is not a prime number because a prime number must have exactly two distinct factors (1 and itself). Since 1 only has one factor, it is neither prime nor composite.

    How can I quickly determine if a large number is divisible by 4?

    A number is divisible by 4 if the last two digits of the number form a number that is divisible by 4. For example, 1,032 is divisible by 4 because 32 is divisible by 4.

    What is the strategy for remainder questions with large powers?

    The best strategy is to look for a pattern in the units digits or the remainders of smaller powers. This cyclical behavior usually repeats every 2 or 4 terms, allowing you to predict the result for any large exponent. You can also use the AI Question Generator to practice more of these pattern-based problems.

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