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# If n is an integer between 1 and 96 (inclusive)-Probability-PS

Author Message
Director
Joined: 08 Jun 2013
Posts: 515
Location: India
GMAT 1: 200 Q1 V1
GPA: 3.82
WE: Engineering (Other)
If n is an integer between 1 and 96 (inclusive)-Probability-PS  [#permalink]

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11 Aug 2018, 08:03
1
00:00

Difficulty:

75% (hard)

Question Stats:

47% (02:10) correct 53% (01:58) wrong based on 15 sessions

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If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?

A) 1/4

B) 1/2

C) 5/8

D) 3/4

E) 7/8

_________________

It seems Kudos button not working correctly with all my posts...

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is it?....

Anyways...Thanks for trying

Senior Manager
Joined: 18 Jun 2018
Posts: 252
If n is an integer between 1 and 96 (inclusive)-Probability-PS  [#permalink]

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11 Aug 2018, 08:40
1
Harshgmat wrote:
If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?

A) 1/4

B) 1/2

C) 5/8

D) 3/4

E) 7/8

OA: C
There are two different possible pattern
Pattern 1: n= odd, then n+1=Even and n+2 = odd i.e odd*even*odd
Number of such arrangement $$= \frac{96}{2}= 48$$
Pattern 2: n= Even, then n+1=Odd and n+2 = Even i.e even*odd*even
Number of such arrangement $$= \frac{96}{2}= 48$$

In Pattern 1 : n+1 should be divisible by 8 as n and n+2 are odd
n+1=8p ; n = 8p-1
Possible value of n = 7,15,23,31,39,47,55,63,71,79,87,95
Total possible values of n = 12
In pattern 2: all 48 are divisible by 8 , as smallest possible case: 2*3*4 is also divisible by 8

Favourable case $$: 12+48 =60$$
Total case :$$96$$
Probability $$= \frac{60}{96} =\frac{5}{8}$$
Intern
Joined: 18 Nov 2017
Posts: 30
Re: If n is an integer between 1 and 96 (inclusive)-Probability-PS  [#permalink]

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11 Aug 2018, 09:23
Bismarck wrote:
Harshgmat wrote:
If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?

A) 1/4

B) 1/2

C) 5/8

D) 3/4

E) 7/8

OA: C
There are two different possible pattern
Pattern 1: n= odd, then n+1=Even and n+2 = odd i.e odd*even*odd
Number of such arrangement $$= \frac{96}{2}= 48$$
Pattern 2: n= Even, then n+1=Odd and n+2 = Even i.e even*odd*even
Number of such arrangement $$= \frac{96}{2}= 48$$

In Pattern 1 : n+1 should be divisible by 8 as n and n+2 are odd
n+1=8p ; n = 8p-1
Possible value of n = 7,15,23,31,39,47,55,63,71,79,87,95
Total possible values of n = 12
In pattern 2: all 48 are divisible by 8 , as smallest possible case: 2*3*4 is also divisible by 8

Favourable case $$: 12+48 =60$$
Total case :$$96$$
Probability $$= \frac{60}{96} =\frac{5}{8}$$

is there any other approach?(faster)
Well explained though.
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: If n is an integer between 1 and 96 (inclusive)-Probability-PS  [#permalink]

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11 Aug 2018, 12:26
Harshgmat wrote:
If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?

A) 1/4

B) 1/2

C) 5/8

D) 3/4

E) 7/8

Discussed here: https://gmatclub.com/forum/if-n-is-an-i ... 60187.html
_________________
Re: If n is an integer between 1 and 96 (inclusive)-Probability-PS &nbs [#permalink] 11 Aug 2018, 12:26
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