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If an integer n is to be chosen at random from the integers 1 to 96 inclusive

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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/
Official Answer

1 Best Answer

Best answer
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.
Answer: D.
answered Sep 2, 2014 by Partner (696 points)
selected Sep 2, 2014 by
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