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    Hard GRE Number Properties Questions Practice Questions

    July 8, 202611 min read13 views
    Hard GRE Number Properties Questions Practice Questions

    Eighty percent of test-takers struggle with the subtle nuances of integer constraints and divisibility rules that define the most challenging quantitative sections. Hard GRE Number Properties Questions Practice Questions focus on the fundamental characteristics of integers, including prime factorization, remainders, and the parity of algebraic expressions. These concepts form the bedrock of the GRE Prep curriculum because they test logic rather than just calculation. By mastering these patterns, you can navigate complex quantitative reasoning problems with greater efficiency. For those seeking more variety, exploring Free GRE Practice Questions Practice Questions with Answers can provide a broader overview of the exam's quantitative landscape.

    Concept Explanation

    Number properties are the inherent rules and characteristics that govern integers, such as their ability to be divided, their status as even or odd, and their prime components. At the advanced level, these questions rarely ask for simple arithmetic; instead, they require you to manipulate variables under specific constraints, such as n n being a positive integer or p p being a prime number greater than 3. Key areas include Divisibility and Remainders, where the relationship Dividend = ( Divisor × Quotient ) + Remainder \text{Dividend} = ( \text{Divisor} \times \text{Quotient}) + \text{Remainder} is frequently tested. Another critical area is Prime Factorization, which is used to determine the number of factors a value has or to find the Greatest Common Divisor (GCD) and Least Common Multiple (LCM). Understanding these properties is essential for succeeding on the Adaptive GRE Practice Test Practice Questions, as the difficulty of these problems scales based on your performance.

    Solved Examples

    1. Example 1: Factorials and Primes
      If n = 15 ! + 2 n = 15! + 2 , is n n divisible by any prime number p p such that 2 ≤ p ≤ 15 2 \leq p \leq 15 ?
      1. Recall that 15 ! = 15 × 14 × 13 × ⋯ × 2 × 1 15! = 15 \times 14 \times 13 \times \dots \times 2 \times 1 . This means 15 ! 15! is divisible by every integer from 2 to 15.
      2. The expression is 15 ! + 2 15! + 2 . Since 2 divides 15 ! 15! and 2 divides 2, then 2 must divide the entire sum ( 15 ! + 2 ) (15! + 2) .
      3. Therefore, n n is divisible by the prime number 2, which falls within the range 2 ≤ p ≤ 15 2 \leq p \leq 15 .
      4. Answer: Yes, it is divisible by 2.
    2. Example 2: Remainder Theory
      When the positive integer n n is divided by 7, the remainder is 3. What is the remainder when 5 n 5n is divided by 7?
      1. Represent n n using the remainder formula: n = 7 k + 3 n = 7k + 3 , where k k is an integer.
      2. Multiply the expression by 5: 5 n = 5 ( 7 k + 3 ) = 35 k + 15 5n = 5(7k + 3) = 35k + 15 .
      3. Rewrite 15 in terms of 7: 15 = ( 2 × 7 ) + 1 15 = (2 \times 7) + 1 .
      4. Substitute back: 5 n = 35 k + 14 + 1 = 7 ( 5 k + 2 ) + 1 5n = 35k + 14 + 1 = 7(5k + 2) + 1 .
      5. The remainder is 1.
    3. Example 3: Parity and Exponents
      If x x and y y are integers and x 2 − y 2 x^2 - y^2 is odd, which of the following must be true: x + y x + y is odd, or x y xy is even?
      1. The expression x 2 − y 2 x^2 - y^2 can be factored as ( x − y ) ( x + y ) (x - y)(x + y) .
      2. For a product to be odd, both factors must be odd. Thus, ( x + y ) (x + y) must be odd.
      3. If ( x + y ) (x + y) is odd, one number must be even and the other must be odd.
      4. If one of x x or y y is even, their product x y xy must be even.
      5. Answer: Both statements must be true.

    Practice Questions

    1. If k k is an integer and 0 < k < 10 0 < k < 10 , for how many values of k k is k ! + 2 k! + 2 divisible by 2?

    2. The product of three consecutive integers is always divisible by which of the following? (Select all that apply: 2, 3, 4, 6).

    3. If x x and y y are positive integers such that x x is a multiple of 6 and y y is a multiple of 15, then x 2 + y 2 x^2 + y^2 must be a multiple of which integer?

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    4. If n n is a positive integer, what is the greatest common divisor of n 2 + n n^2 + n and n + 1 n + 1 ?

    5. Let p p be a prime number greater than 3. What is the remainder when p 2 p^2 is divided by 12?

    6. If m m is an even integer and n n is an odd integer, which of the following must be an odd integer? ( m + 1 ) 2 + n 2 (m+1)^2 + n^2 , m 2 + ( n − 1 ) 2 m^2 + (n-1)^2 , or m n + 1 mn + 1 ?

    7. How many trailing zeros are in the decimal representation of 25 ! 25! ?

    8. If a , b , c a, b, c are distinct prime numbers, how many factors does the number a 2 b c 3 a^2 b c^3 have?

    9. A number X X when divided by 12 leaves a remainder of 5. What is the remainder when X 2 X^2 is divided by 8?

    10. If x x is a multiple of 9 and y y is a multiple of 12, what is the largest integer that must divide x + y x + y ?

    Answers & Explanations

    1. 9 values: For any k ≥ 2 k \geq 2 , k ! k! is even because it contains the factor 2. Since 2 is also even, the sum k ! + 2 k! + 2 is always even (divisible by 2) for k = 2 , 3 , … , 9 k = 2, 3, \dots, 9 . For k = 1 k = 1 , 1 ! + 2 = 3 1! + 2 = 3 , which is not divisible by 2. Thus, there are 8 values. Wait—rechecking the range 0 < k < 10 0 < k < 10 , the integers are 1 through 9. Only k = 1 k=1 fails. So 8 values.
    2. 2, 3, and 6: In any three consecutive integers, at least one is even (divisible by 2) and exactly one is a multiple of 3. Therefore, the product must be divisible by 2 × 3 = 6 2 \times 3 = 6 . It is not always divisible by 4 (e.g., 1 × 2 × 3 = 6 1 \times 2 \times 3 = 6 ).
    3. 9: Let x = 6 a x = 6a and y = 15 b y = 15b . Then x 2 = 36 a 2 x^2 = 36a^2 and y 2 = 225 b 2 y^2 = 225b^2 . The Greatest Common Divisor of 36 and 225 is 9. Thus, 36 a 2 + 225 b 2 = 9 ( 4 a 2 + 25 b 2 ) 36a^2 + 225b^2 = 9(4a^2 + 25b^2) .
    4. n + 1: Factor the first expression: n 2 + n = n ( n + 1 ) n^2 + n = n(n + 1) . The two expressions are n ( n + 1 ) n(n + 1) and ( n + 1 ) (n + 1) . The largest term that divides both is n + 1 n + 1 .
    5. 1: Any prime p > 3 p > 3 can be written as 6 k ± 1 6k \pm 1 . Then p 2 = ( 6 k ± 1 ) 2 = 36 k 2 ± 12 k + 1 p^2 = (6k \pm 1)^2 = 36k^2 \pm 12k + 1 . Factoring out 12 gives 12 ( 3 k 2 ± k ) + 1 12(3k^2 \pm k) + 1 . The remainder is always 1.
    6. mn + 1: If m m is even and n n is odd, m n mn is even. Adding 1 to an even number results in an odd number. The other options involve sums of squares that result in even numbers: ( odd ) 2 + ( odd ) 2 = even ( \text{odd})^2 + ( \text{odd})^2 = \text{even} .
    7. 6: Trailing zeros are created by pairs of factors 2 and 5. In a factorial, 5s are rarer. Count multiples of 5 in 25 ! 25! : 5, 10, 15, 20, 25. Note that 25 has two factors of 5 ( 5 × 5 5 \times 5 ). Total 5s = 1 + 1 + 1 + 1 + 2 = 6 1 + 1 + 1 + 1 + 2 = 6 .
    8. 24: For a number p 1 a p 2 b p 3 c p_1^a p_2^b p_3^c , the number of factors is ( a + 1 ) ( b + 1 ) ( c + 1 ) (a+1)(b+1)(c+1) . Here, ( 2 + 1 ) ( 1 + 1 ) ( 3 + 1 ) = 3 × 2 × 4 = 24 (2+1)(1+1)(3+1) = 3 \times 2 \times 4 = 24 .
    9. 1: Let X = 12 k + 5 X = 12k + 5 . Then X 2 = ( 12 k + 5 ) 2 = 144 k 2 + 120 k + 25 X^2 = (12k + 5)^2 = 144k^2 + 120k + 25 . Both 144 and 120 are divisible by 8. Dividing 25 by 8 gives a remainder of 1 ( 8 × 3 = 24 8 \times 3 = 24 ).
    10. 3: Since x = 9 a x = 9a and y = 12 b y = 12b , x + y = 3 ( 3 a + 4 b ) x + y = 3(3a + 4b) . The GCD of 9 and 12 is 3. We cannot guarantee a larger factor because 3 a + 4 b 3a + 4b could be any integer depending on a a and b b .
    Interactive quizQuestion 1 of 5

    1. If \( n \) is an even integer, which of the following must be even?

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    Frequently Asked Questions

    What are the most common number properties tested on the GRE?

    The GRE frequently tests concepts such as odd and even parity, divisibility rules for 3, 4, and 9, prime factorization, and remainder arithmetic. Understanding these fundamentals allows students to solve complex algebraic puzzles more quickly.

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

    To find the total number of factors, prime factorize the number into the form p 1 a p 2 b … p_1^a p_2^b \dots , then add one to each exponent and multiply the results. For example, 12 = 2 2 × 3 1 12 = 2^2 \times 3^1 has ( 2 + 1 ) ( 1 + 1 ) = 6 (2+1)(1+1) = 6 factors.

    Is 1 a prime number on the GRE?

    No, by definition, a prime number must be an integer greater than 1 that has exactly two divisors. Therefore, 2 is the smallest prime number and the only even prime number.

    What is the difference between a factor and a multiple?

    A factor is a number that divides into another number evenly without a remainder, while a multiple is the product of a given number and an integer. For instance, 4 is a factor of 12, and 12 is a multiple of 4.

    How can I quickly determine if a large number is divisible by 3?

    A number is divisible by 3 if the sum of its digits is divisible by 3. This property, along with similar rules for 9, is a staple of standardized test number theory.

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