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Tough and tricky: Probability of integr being divisible by 8

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

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11 Oct 2009, 18:33
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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Oct 2009, 18:39
There could be 36 such integers. So probability according to me should be 36/96 = 37.5%.
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Oct 2009, 18:54
hgp2k wrote:
There could be 36 such integers. So probability according to me should be 36/96 = 37.5%.

Checked answered choices, then solved again myself, seems choices are correct. Can you please share your way of thinking?
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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%

Please correct me if I am wrong.
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Oct 2009, 22:18
Hmm. isnt it C - 62.5

Here is my logic behind it..

The product is a combination of 3 numbers. There a total of 96 of these.

Any combination with two even numbers (starting with n=2) can be divided by 8. - 48 possible outcomes

Of the combinations with two even numbers the ones with multiples of 8 (eg: when n=7,15,23) are also divisible by 8. - 12 possible outcomes.

60/96 = 62.5%
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Oct 2009, 22:39
I ll go with C. Whats the OA ???
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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12 Oct 2009, 04:11
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If n is odd then only 12 choices - 7,15,23 etc....

if n is even then for all n it is divisible by 8
as n*(n+2) is divisible by 8 for all even n

Hence total numbers divisible by 8 = 48+12 = 60

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

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12 Oct 2009, 05:32
hmmmm, I got it now. I was considering only the multiples of 8 and not all even numbers.
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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12 Oct 2009, 09:29
Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
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Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]

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11 Feb 2011, 05:39
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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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Re: Tough and tricky: Probability of integr being divisible by 8   [#permalink] 11 Feb 2011, 05:39
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