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If n is a positive integer, then n(n+1)(n+2) is (A) even on
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17 Aug 2020, 11:15
17 Aug 2020, 11:15
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If n is a positive integer, then n(n+1)(n+2) is
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
Re: If n is a positive integer, then n(n+1)(n+2) is (A) even on
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17 Aug 2020, 11:16
17 Aug 2020, 11:16
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Re: If n is a positive integer, then n(n+1)(n+2) is (A) even on
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17 Aug 2020, 11:28
17 Aug 2020, 11:28
Carcass wrote:
If n is a positive integer, then n(n+1)(n+2) is
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
(A) even only when n is even
(B) even only when n is odd
(C) odd whenever n is odd
(D) divisible by 3 only when n is odd
(E) divisible by 4 whenever n is even
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.
Notice that n, n+1, and n+2 are three consecutive integers.
This means the product of n, n+1, and n+2 will be divisible by 3, 2 and 1
Since n(n+1)(n+2) is divisible by 2, then the product is ALWAYS EVEN.
This means we can eliminate answer choices A and B, because they put restrictions on when the product is even.
We can also eliminate C because it suggests that the product can be odd.
Likewise, since n(n+1)(n+2) is ALWAYS divisible by 3, we can eliminate answer choice D, because it puts a restriction on when the product is divisible by 3
Answer: E
Cheers,
Brent
