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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