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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}
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Jul 2013, 05:23, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Easy probability [#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: Easy probability [#permalink] New post 29 Dec 2009, 08:28
i am sorry, the options were wrong, have edited it :)
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Re: Easy probability [#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: Easy probability [#permalink] New post 29 Dec 2009, 08:50
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. 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: Easy probability [#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: Easy probability [#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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Numbers [#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: Numbers [#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: Numbers [#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

D
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Re: Easy probability [#permalink] New post 02 May 2010, 13:46
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Re: Easy probability [#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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Number Properties [#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: Number Properties [#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: Number Properties [#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: Number Properties [#permalink] New post 17 May 2010, 14:30
good problem great explanation.
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Re: Easy probability [#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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Probability Question [#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.
Probability Question   [#permalink] 08 Sep 2010, 10:18
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