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

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29 Dec 2009, 08:13

kirankp 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. 41 B. 83 C. 21 D. 85 E. 43

are you sure the options are correct?? ...as per my understanding all even numbers in this range 2 4 6 8 till 96 will be divisible by 8 for n(n+1)(n+2)....what am I missing here

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. 41 B. 83 C. 21 D. 85 E. 43

\(n(n+1)(n+2)\) 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)=48

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.

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

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02 Jan 2010, 02:53

agree with Bunuel here. if n=even, then n(n+2) will always have at least three 2's as factors (2x2x2=8), therefore 8 will always be a factor, where n= even. there are 48.

if (n+1)= divisible by 8, then the whole equation will also be divisible by 8: n=7,15,23,31,39,47,55,63,71,79,87,95 = 12 possibilities.

48+12=60 -- number of times expression will be divisible by 8. 96= number of possibilities.

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

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02 May 2010, 10:47

gurpreetsingh wrote:

chaitu1315 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

IMO A

This is only possible when n is multiple of 4.

now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4

Gurpreet you did not consider the options mentioned below ..!

What about when n =6 or may be 15 then also the equation is divisible by 8 and hence we need to consider them as well..

so in my opinion the expression will be divisible by 8 whenever n is divisible by 4 or (n+ 2) is divisible by 4 or when (n+1) is divisible by 8

1st Option when n is divisible by = total numbers = 24 ( as calculated by Gurpreet above 96/4= 24) 2nd option when n +1 is divisible by 8 = 12 ( 96/8 ) 3rd Option when n+2 is divisible by 4 = 24 ( 96/4)

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

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02 May 2010, 19:45

1

This post received KUDOS

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This post was BOOKMARKED

kirankp 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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D.\(\frac{5}{8}\) E. \(\frac{3}{4}\)

If you notice every alternate pair is divisible by 8 => 48 such pairs If you notice the pair(marked in blue) it has '8' in it and hence divisible by '8'=>12 such pairs

=>prob=48+12 / 96 = 5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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11 May 2010, 09:15

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

The problem can be solved in a crude way, but is there a better alternative?

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

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11 May 2010, 19:09

ajitsah 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

The problem can be solved in a crude way, but is there a better alternative?

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

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11 Aug 2010, 05:36

I am curious... how should we approach this type of question...? for eg this question says 8 if question says 11 or may be 12 do we test numbers? what if there are certain exception in middle? (in case of 8 it seems it goes continously...)

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

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08 Sep 2010, 10:18

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. 3/8

Last edited by udaymathapati on 09 Sep 2010, 04:47, edited 1 time in total.

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

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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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{5}{8}\) E. \(\frac{3}{4}\)

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

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8;

(Notice 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.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

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