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What is the probability of n(n+1)(n+2) being evenly divisibl

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What is the probability of n(n+1)(n+2) being evenly divisibl

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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: www.GMATinsight.com

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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: 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
\[\frac{{\# \left\{ {n\left( {n + 1} \right)\left( {n + 2} \right)\,\,{\text{divisible}}\,\,{\text{by}}\,\,{\text{8}}:n\,\,\,\operatorname{int} } \right\}}}{{\# \left\{ {n\left( {n + 1} \right)\left( {n + 2} \right):n\,\,\,\operatorname{int} } \right\}}}\]
DOES have a precise mathematical meaning and a unique value... the reasoning to obtain it rigorously is far out-of-GMAT`s scope.

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

---------------------------------------------------------------------------------------------------
n=1 implies n(n+1)(n+2) is not divisible by 8
n=2 implies n(n+1)(n+2) IS divisible by 8
n=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
---
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.
----------------------------------------------------------------------------------------------------------
(*) The proper way to formalize this reasoning is the real issue here... we have an infinite number of blocks to
evaluate, and although it seems that things must go in exactly the same way... this is not as easy as it seems!


(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" problem!)

> 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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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
Statement 1:
Since n is an integer, n(n+1)(n+2) = the product of 3 consecutive integers.
WRITE IT OUT and LOOK FOR A PATTERN.

1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10


9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18


Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.

Thus, the probability that n(n+1)(n+2) will be a multiple of 8 = 5/8.
SUFFICIENT.

Statement 2:
Since it is unknown whether n is an integer, we cannot determine the probability that n(n+1)(n+2) will be a multiple of 8.
INSUFFICIENT.

The correct answer is A.

An alternate approach for Statement 1:
If n is an integer, then n(n+1)(n+2) will be a multiple of 8 in two cases.

Case 1: n is even, with the result that n(n+1)(n+2) = even*odd*even
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
P(n is an even integer) = 1/2.

Case 2: n+1 is a multiple of 8
n(n+1)(n+2) will be a multiple of 8 if n+1 is a multiple of 8.
Since one of every 8 integers is a multiple of 8, we get:
P(n+1 is a multiple of 8) = 1/8.

Since a favorable outcome will be yielded by Case 1 or Case 2, we ADD the probabilities:
1/2 + 1/8 = 4/8 + 1/8 = 5/8.

The problem above is modeled after the following problem in GMATPrep:
https://www.beatthegmat.com/probability-of-n-n-1-n-2-divisible-by-8-t281883.html
For a similar problem, check here:
https://www.beatthegmat.com/probability-t116280.html

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