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Re: If an integer n is to be chosen at random from the integers [#permalink]
11 Nov 2007, 20:22

2

This post was BOOKMARKED

plaza202 wrote:

If an integer n is to be chosen at randon 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.

please help .................

I will go with D 5/8

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?

total nos of ways in which we can choose n = 96

n(n + 1)(n + 2) will be divisible by 8?

case 1: n = odd then n+2 =odd & n+1 will be even i.e this needs get divided by 8, hence is a multiple of 8 so we have 8..96 = 12 multiples to fill the n+1 pos hence 12 ways

case 2: n is even then n+2 will be even & the product will be divisible by 24 & thus 8

so nos of values that can be used for n= 2....96 (all even nos) i.e 48 nos

Re: If an integer n is to be chosen at random from the integers [#permalink]
12 Nov 2007, 09:57

The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.

ex. 2,3,4

The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).

There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.

Re: If an integer n is to be chosen at random from the integers [#permalink]
12 Nov 2007, 10:01

gixxer1000 wrote:

The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.

ex. 2,3,4

The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).

There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.

Ans C

What abt 7,8,9

starts with an odd integer but is still divisible by 8

Re: If an integer n is to be chosen at random from the integers [#permalink]
26 Aug 2008, 13:27

5/8 should be the answer. 48 numbers are even for n 12 triplets have multiples of 8 in them --- this implies that those odd numbers that are in those triplets are all game, e.g., 7-8-9 ,15-16-17, 23-24-25 etc. Therefore, 7, 15, 23 would also be counted as possible n. There are 12 such nos. Therefore, the total should be 48 +12 = 60. Hence, prob. = 60/96 = 5/8

Re: If an integer n is to be chosen at random from the integers [#permalink]
26 Jul 2014, 05:02

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If an integer n is to be chosen at random from the integers [#permalink]
26 Jul 2014, 08:25

1

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Expert's post

5

This post was BOOKMARKED

plaza202 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:

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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