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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] New post 29 Dec 2009, 04:41
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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. \(\frac{3}{4}\)

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Apr 2015, 07:16, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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 :? :x
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 29 Dec 2009, 08:28
i am sorry, the options were wrong, have edited it :)
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 29 Dec 2009, 08:31
kirankp wrote:
i am sorry, the options were wrong, have edited it :)


hmmm...will go with C-1/2
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 29 Dec 2009, 08:50
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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. 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.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 29 Dec 2009, 08:52
Bunuel wrote:
n(n+1)(n+2) is divisible by 8 in two cases:

1. When n is even:
No of even numbers (between 1 and 96)=48

AND

2. When n+1 is divisible by 8. --> n+1=8p, n=8p-1 --> 8p-1<=96 --> p<=12.1 --> 12 such numbers

Also note that these two sets have no overlaps.

Total=48+12=60

Probability=60/96=5/8

Answer: D.


oops :cry: :oops:
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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.

probability = 60/96 = 5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 02 May 2010, 08:16
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:
D

Last edited by chaitu1315 on 02 May 2010, 08:24, edited 1 time in total.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 02 May 2010, 08:22
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
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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)

in total = 24 + 24 + 12 = 60

the probability = 60/ 96 = 5/ 8

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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 02 May 2010, 13:46
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 02 May 2010, 19:45
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. \(\frac{3}{4}\)



IMO 'D'
Consider the following possibilities:
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
.......
......
96 97 98

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] New post 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?
[Reveal] Spoiler:
OA : D


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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 11 May 2010, 18:52
I do not know how to solve this. Where did you get this problem?
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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?
[Reveal] Spoiler:
OA : D


Thanks


Probability = required cases/ total cases.

required cases . Case 1. n(n + 1)(n + 2) will be divisible by 8 when n is even, in that case, n will be multiple of 2 and n+2 of 4.

so total even numbers between 1 and 96 = 48

Case 2. when n+1 = 8k form => n = 8k-1

such numbers from 1 to 96 are 12... it starts from 7 to 95 which diff of 8.This forms an AP series.

So required cases = 48+12 = 60

total cases = 96

probability = 60/96 = 5/8

Hence D
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 17 May 2010, 14:30
good problem great explanation.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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...)

do we remember for each numbers?
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 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.
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 20 Apr 2015, 06:07
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 20 Apr 2015, 07:16
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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. \(\frac{3}{4}\)


\(n(n + 1)(n + 2)\) 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.

Answer: D.

Similar question: divisible-by-12-probability-121561.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 20 Apr 2015, 07:16
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