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Probabilty and Divisor

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by Kaunteya » Sat Mar 01, 2008 10:33 am
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

I get 1/2. Can some comment on how to tackle this question. The right answer is D. I don't see it. I figured out from 0-9 which value using the formula above which are divisible by 8. I discoved that 2, 4, 6, 7, 8 are divisible by 8. If 2 is divisible by 8 the same should go for 12, 22, 32 and so on. Counting them up I got 48 numbers divisible by 8. 48/96 = 1/2. Where did I go wrong?

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by iwillmakeit » Sat Mar 01, 2008 9:34 pm
D

take a look at integer sequence:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19

to catch 8*n we have to start with 8n-2, 8n-1, 8n

1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19

Therefore, we have 3 combinations for each 8 consecutive integers.

to catch 8*n we also have to start with other even integers 8n-6, 8n-4

1,[2,3,4],5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
1,2,3,[4,5,6],7,8,9,10,11,12,13,14,15,16,17,18,19

p=3/8+ 2/8=5/8
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