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The product of n consecutive positive integers is always divisible by
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25 Feb 2022, 07:49
25 Feb 2022, 07:49
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The product of n consecutive positive integers is always divisible by
(A) \(n^2 - 1\)
(B) \((n + 1)!\)
(C) \(2n + 1\)
(D) \(n^2 + 1\)
(E) \(n!\)
(A) \(n^2 - 1\)
(B) \((n + 1)!\)
(C) \(2n + 1\)
(D) \(n^2 + 1\)
(E) \(n!\)
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Re: The product of n consecutive positive integers is always divisible by
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25 Feb 2022, 09:41
25 Feb 2022, 09:41
Carcass wrote:
The product of n consecutive positive integers is always divisible by
(A) \(n^2 - 1\)
(B) \((n + 1)!\)
(C) \(2n + 1\)
(D) \(n^2 + 1\)
(E) \(n!\)
(A) \(n^2 - 1\)
(B) \((n + 1)!\)
(C) \(2n + 1\)
(D) \(n^2 + 1\)
(E) \(n!\)
APPROACH #1: Apply a property of consecutive integers
There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.
So, the product of n integers must be divisible by n, n-1, n-2, ...2, 1
This means the product of n integers must be divisible by n!
Answer: E
APPROACH #2: Test values
Let's test n = 2
So, for example, the product of 2 integers could = (2)(1) = 2
For each answer choice, we'll plug n = 2 enter the expression and check whether it's a divisor of 2
(A) 2^2 - 1 = 3. Since 2 is NOT divisible by 3, we can eliminate choice A.
(B) (2 + 1)! = 3. Since 2 is NOT divisible by 3, we can eliminate choice B.
(C) 2(2) + 1 = 5. Since 2 is NOT divisible by 5, we can eliminate choice C.
(D) 2^2 + 1 = 5. Since 2 is NOT divisible by 5, we can eliminate choice D.
(E) 2! = 2. Since 2 is divisible by 2, we'll KEEP choice E.
Answer: E
gmatclubot
Re: The product of n consecutive positive integers is always divisible by [#permalink]
25 Feb 2022, 09:41
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