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If n is the smallest of three consecutive positive integers

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If n is the smallest of three consecutive positive integers

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If n is the smallest of three consecutive positive integers, which of the following must be true?

(A) n is divisible by 3
(B) n is even
(C) n is odd
(D) (n)(n + 2) is even
(E) n(n + 1)(n + 2) is divisible by 3

OA=E

Source: Manhattan GMAT

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VJesus12 wrote:
If n is the smallest of three consecutive positive integers, which of the following must be true?

(A) n is divisible by 3
(B) n is even
(C) n is odd
(D) (n)(n + 2) is even
(E) n(n + 1)(n + 2) is divisible by 3
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

By the above rule, the product of 3 consecutive integers will be divisible by 3, 2 and 1
If we recognize that n, n+1 and n+2 represent 3 consecutive integers
So, n(n + 1)(n + 2) must be divisible by 3

Answer: E

Cheers,
Brent

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The product of 3 consecutive positive integers is divisible by 6

\(6=3*26=3-2\), Thus the Product of \(3\) consecutive positive integers is divisible by both \(2\) and \(3\), Answer must be (E)

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VJesus12 wrote:
If n is the smallest of three consecutive positive integers, which of the following must be true?

(A) n is divisible by 3
(B) n is even
(C) n is odd
(D) (n)(n + 2) is even
(E) n(n + 1)(n + 2) is divisible by 3

OA=E

Source: Manhattan GMAT
Looking at the answer choices, we see that answer choice E represents the product of 3 consecutive integers, and that product is always divisible by 3! = 6, so it also must be divisible by 3.

Answer: E

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scott@targettestprep.com



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