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

\(? = P\left( {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8} = \operatorname{int} } \right)\)

(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 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
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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.

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

1) n is an integer
2) 1≤ n ≤ 96

Why option B is wrong because I am getting answer as 1/4 from option B.
And pls someone explain how we can get the probability using option A.

Thanks in advance
Anurag Jain

jainanurag470

To define a probability you need a sample space or the number of all possible outcomes.
Statemetn 2 does not define if n is integer or not, hence we can not count the total no of possible cases.

Thank you.
But I still doubt option A because sample space is not clear in option A too.
for eg. If I take first 8 positive intgers as my sample space then probability will be = 3/8
And if I take first 9 positive intgers as my sample space then probability will be = 3/9=1/3

Thanks in advance
Anurag Jain

Hi, Anurag!

In the question stem pre-statements combined with statement (1), the sample space is the whole Z, in other words, the set of integers.

We are dealing with the product of ANY three consecutive integers, including negative ones, by the way!

As I mentioned, the proper way of calculating the probability asked is NOT in GMAT´s scope, but the "informal argument"
(interesting and PERFECT when (1+2) is considered) was presented in my solution.

Regards,
fskilnik.

Hi fskilink,

First of all thank you for the solution.
I am convinced that solution is C but OA is A.
And with your provided solution I am not convinced that solution is A.

Thanks in advance
Anurag Jain

Dear Anurag,

Thank you for your contact and your strong desire to understand. That´s exactly what makes us stronger/wiser!

First of all, I believe I understand the reason you believe the answer is (C). You think something like this:

(1) If 1 <= N <= 2 , we have one answer (1/2) , but if 1 <= N <= 3 we have another answer (1/3), correct?

The rationale IS correct, but we must understand the statement (1) without being "contaminated" by the idea presented in (2)...

More explicitly: we look for the cases in which the expression n(n+1)(n+2) is divisible by 8, and using statement (1) only, we just know that n is an integer. Hence our question turns out to be:

Is n(n+1)(n+2) divisible by 8, when n is an integer? (Only considering the question stem pre-statements and statement (1) , both combined.)

Now I guess you understand that the expression n(n+1)(n+2) "runs" over infinite possibilities, because we must analyse ALL cases involved, some of them are the following:

when n = -200, we have to evaluate -200*(-199)*(-198)
when n = 0, we have to evaluate 0*1*2
when n = 1029 we have to evaluate 1029*1030*1031

In other words, we must ask ourselves the following: for EACH integer n, putting the value of the product n(n+1)(n+2) written in a small paper inside a box, (three small papers are with the numbers shown in red above) , do you understand we have infinite small papers inside the box?
(I didn´t say all numbers are different, because (-2)(-1)(0) equals (-1)(0)(1) for instance!)

If you understand that, great, because now you understand the question in blue: what is the probability that randomly choosing one small paper, you will get a number divisible by 8?

The answer IS possible to obtain, and it will be 5/8. The problem is to justify that using formal math. But, as I mentioned, this is out-of-our-scope!

I hope now things are clear!

Regards,
fskilnik.

P.S.: try my test-drive, please! Reason: students who want to understand things deeply and "as sub-product" to perform in REALLY higher-level are EXACTLY GMATH method profile...

Does the term 'evenly divisible' apply for negative integers? Or should option A specify that n is a "positive integer or 0"?

Hi AyuPathak

1) Evenly Divisible mean no remainder when divided by a divisor. This term is also applicable for negative integers such as -2 is evenly divisible by 2
However, we use the terms multiples and factors only in context of positive integers

2) This question does not require n to be positive because even for negative values, probability of product of three consecutive integer is always divisible by 5/8

how can you take consecutive integers to get the probability 5/8? It is said n as integer. i can take 1 and 17 or any other number....Why consecutive integers are not taking in case II? please help me to understand it.....