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1) if n is even, n(n+1)(n+2) is necessarily a multiple of 8 when n>=2

n = 2K:
= 2K(2K+1)(2K+2)
= 4K(2K+1)(K+1)

Using the above, all I can say is it is a multiple of 4.

Let us proceed further..

K even: If K is even, we can tell that above is a multiple of 8.

K Odd: If K is odd = 2P-1
=4(2P-1)(2P-2 +1) (2P-1 +1)
=4(2P-1)(2P-1)(2p)
Aha ... it is a multiple of 8!
Proved!

2) if n is odd, then n(n+1)(n+2) is a multiple of 8 if only (n+1) is.

n = 2K-1:
=(2K-1)(2K)(2K+1)
=2(2K-1)(K)(2K+1)

Using the above, all I can say is it is a multiple of 2.

Although, we can say that the above product is a multiple of 8 if n is odd only if n+1 is a multiple of 8 looking at the product as n and n+2 will still be odd.
Proved.

Re: If an integer n is to be chosen at random from the integers [#permalink]

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17 Oct 2013, 13:46

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If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4

\(n(n + 1)(n + 2)\) is divisible by 8 in two cases:

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8;

(Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.