GMATinsight wrote:

What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer

2) 1≤ n ≤ 96

Source:

http://www.GMATinsight.com\(? = P\left( {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8} = \operatorname{int} } \right)\)

(1) Although the ratio we are asked to obtain

\(\frac{{\,\,\,\# \,\,\left\{ {\,\,n\left( {n + 1} \right)\left( {n + 2} \right){\mkern 1mu} {\mkern 1mu} \,\,{\text{divisible}}{\mkern 1mu} {\mkern 1mu} {\text{by}}{\mkern 1mu} {\mkern 1mu} {\text{8}}\,\,\,:\,\,\,n{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \operatorname{int} \,\,\,} \right\}\,\,\,}}{{\# \,\,\left\{ {\,\,\,n\left( {n + 1} \right)\left( {n + 2} \right)\,\,\,:\,\,\,n\,\,{\mkern 1mu} \operatorname{int} \,\,\,} \right\}}}\)

DOES have a precise mathematical meaning and a unique value... the reasoning to obtain it rigorously is out-of-GMAT`s scope.

Anyway, the rationale the examiner´s (probably) had in mind is important:

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n=1 implies n(n+1)(n+2) is not divisible by 8

n=2 implies n(n+1)(n+2) IS divisible by 8n=3 implies n(n+1)(n+2) is not divisible by 8

n=4 implies n(n+1)(n+2) IS divisible by 8 n=5 implies n(n+1)(n+2) is not divisible by 8

n=6 implies n(n+1)(n+2) IS divisible by 8 n=7 implies n(n+1)(n+2) IS divisible by 8 n=8 implies n(n+1)(n+2) IS divisible by 8 n=9 is the beginning of a new cycle, identical to the n=1 situation

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n=16 is the end of this new cycle, identical to the n=8 situation

Hence in every cycle we have

5 favorable cases among 8 equiprobable possibilities (*) , hence the answer is 5/8.

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(*) The real issue is to formalize this reasoning...

we have an infinite number of blocks to evaluate, and

although things "behave" in the same way... the justification is far out-of-GMAT´s universe!

(2) We cannot use the fact that n is an integer here... that´s where we are able to BIFURCATE:

> If n is an integer, the argument shown above (between the parallel lines) is PERFECT and the answer would be 5/8

(The fact that we are dealing with integers in the interval [1,96] avoids the "infinite blocks" higher-level analysis!)

> If n is not necessarily an integer, we could choose n in the interval [1,96] such that n=1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, ... , 95, 95.5, 96 (for instance).

In this case, it is not 5 favorable cases for every 8 (most n´s will give n(n+1)(n+2) not an integer, thefore the "no divisibility by 8" is more frequent)!

This solution follows the notations and rationale taught in the GMATH method.

Regards,

fskilnik.

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Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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