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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Oct 2009, 19:08

Sure. Here is what I think, and I might be wrong (In fact most of the times I am wrong!):

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8. Now, there are 96/8 = 12 integers from 1 to 96 which are divisible by 8. For n*(n+1)*(n+2) to be divisible by 8: n, (n+1) or (n+2) must be divisible by 8. There could be 12*3 = 36 such integers. So the probability for n*(n+1)*(n+2) being divisible by 8 = 36/96. = 3/8 = 37.5%

Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Feb 2011, 05:39

2

This post received KUDOS

this questions tests the concepts of consecutive integers.

three consecutive integers will .. 1. have an integer that is a multiple of 3 e.g. {2,3,4} or {3,4,5} 2. either have two odds or two even integers e.g. {2,3,4} or {3,4,5} 3. have two even integers if n (1st integer) is even. the product n*(n+1)*(n+2) must be a multiple of 8 because one even integer will be a multiple of 4. e.g. {2,4,5}, {4,5,6} or {14,15,16} 4. have two odd integers if n (1st integer) is odd. the product can be a multiple of 8 only if (n+1) is a multiple of 8 because the other two are odd. e.g. {7,8,9} or {23,24,25}. on the other hand, the product of {3,4,5} or {11,12,13} is not multiple of 8 because (n+1) is not multiple of 8.

we can use the above rules to calculate probability.

no. of cases where n is even = \(\frac{96}{2} = 48\) no. of cases where n+1 is multiple of 8 = \(\frac{96}{8} = 12\) total cases in which product is multiple of 8 = \(48+12 = 60\)

probability = \(\frac{60}{96}\) = 62.5%

Ans: C

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