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# If an integer n is to be chosen at random from the integers

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If an integer n is to be chosen at random from the integers [#permalink]

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27 Sep 2009, 03:01
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95% (hard)

Question Stats:

40% (02:30) correct 60% (01:33) wrong based on 121 sessions

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2012, 06:30, edited 1 time in total.
Edited the question.
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14 Feb 2010, 13:07
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marcodonzelli wrote:
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

All even numbers would be divisible by 8... as we have two consecutive even numbers being multiplied... Hence 96-1/2 + 1 = 48

Also we have 8n-1 sequence numbers divisible by 8... i.e, if n =7,15,23,31....

Therefore 8n-1 = 96 to find the number of elements in this sequence ... gives n = 97/8 = 12

Therefore Probability = (48+12) / 96 = 60 / 96 = 5/8
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18 Sep 2010, 21:53
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Expert's post
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

$$n(n+1)(n+2)$$ is divisible by 8 in two cases:

1. When $$n$$ is even:
$$n=2k$$ --> $$n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)$$ --> either $$k$$ or $$k+1$$ is even so 8 is a multiple of $$n(n+1)(n+2)$$.

# of even numbers between 1 and 96, inclusive is $$\frac{96-2}{2}+1=48$$ (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

2. When $$n+1$$ is divisible by 8. --> $$n+1=8p$$ ($$p\geq{1}$$), $$n=8p-1$$ --> $$8p-1\leq{96}$$ --> $$p\leq{12.1}$$ --> 12 such numbers.

Also note that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.

Total=48+12=60

Probability: $$\frac{60}{96}=\frac{5}{8}$$

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19 Sep 2010, 16:31
This was difficult. I was thinking either N can be multiple of 8 or n+1 can be multiple of 8 pr n+2 a multiple 8. With this I just got 36 choices. I missed sequences like 2*3*4
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19 Sep 2010, 16:41
Notice what happens when you shift this product by 8 :

$$(n+8)(n+1+8)(n+2+8) = 8^3 + 8^2*(n+n+1+n+2) + 8*(n*(n+1)+(n+1)*(n+2)+n*(n+2)) + n*(n+1)*(n+2) = 8*K + n*(n+1)*(n+2)$$

Here K is a constant. Therefore, when we shift by 8, the remainder when divided by 8 remains the same !

So if we figure out what happens for the first 8 choices of n, for all the rest the pattern will repeat

n=1 1x2x3 No
n=2 2x3x4 Yes
n=3 3x4x5 No
n=4 4x5x6 Yes
n=5 5x6x7 No
n=6 6x7x8 Yes
n=7 7x8x9 Yes
n=8 8x9x10 Yes

So for every 8 numbers, exactly 5 will be divisible by 8.

Here there are exactly 96 numbers to consider starting with 1 ... So probability will be exactly 5/8
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21 Sep 2010, 06:24
Bunuel wrote:
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

$$n(n+1)(n+2)$$ is divisible by 8 in two cases:

1. When $$n$$ is even:
$$n=2k$$ --> $$n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)$$ --> either $$k$$ or $$k+1$$ is even so 8 is a multiple of $$n(n+1)(n+2)$$.

# of even numbers between 1 and 96, inclusive is $$\frac{96-2}{2}+1=48$$ (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

2. When $$n+1$$ is divisible by 8. --> $$n+1=8p$$ ($$p\geq{1}$$), $$n=8p-1$$ --> $$8p-1\leq{96}$$ --> $$p\leq{12.1}$$ --> 12 such numbers.

Also note that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.

Total=48+12=60

Probability: $$\frac{60}{96}=\frac{5}{8}$$

Thanks for such a good explanation !!!

I was stuck in this for more then 3 minutes and couldn't find the solution.
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09 Oct 2010, 22:29
Thanks Bunuel for a very nice explanation.

I was stuck with this question for a very long time and could not find the solution to it.

Even tried counting but that too went wrong.

Request you to advice for how to tackle such kind of questions and how you spotted there would be two cases to test for this question ?

Regards,
Sachin
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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01 Nov 2013, 05:45
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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17 Oct 2014, 08:23
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

What are the cases when $$n(n+1)(n+2)$$ is divisible by 8?

This means that $$n(n+1)(n+2)$$ must have 2*2*2 in its prime factorization.

If n is even, we'll have 2 even numbers in $$n(n+1)(n+2)$$, one of which is divisible by 4. This means that $$n(n+1)(n+2)$$ is divisible by 8. Therefore, our probability is at least 1/2. We can eliminate choices A,B, and C immediately.

The other possibility is if we have a multiple of 8 in the middle. This means that n will have to be odd, so we're not considering overlapping cases. This happens exactly 1/8 of the time, or (96/8) = 12 times.

1/2 + 1/8 = 5/8.

Re: If an integer n is to be chosen at random from the integers   [#permalink] 17 Oct 2014, 08:23
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