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## Probability n(n+1)(n+2)

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

Not sure, what's wrong with my approach, please guide :

96/8 = 12

P(reqd condition) = P(n is multiple of 8) + p((n+1) is multiple of 8) + P((n+2)) is multiple of 8)

I.e 3 * 12/96

= 3/8

2. Good post? |
If n is multiple of 4 even than it is divisible by 8

96/4 = 24

3* 24/96 = 3/4

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lesfeaves --Am i getting you rite ? do you mean any multiple of 4 divisible by 8 ????

4. Good post? |

If you pick any even number n from 1-96, then n*(n+1)*(n+2) will be divisible by 8

so 96/2=48 even numbers

Also, there are 12 numbers divisible by 8, 96/8=12

So you can pick 12 odd numbers that are 1 less than a mulltiple of 8 (7,15,23...95) and you will get n*(n+1)*(n+2) divisible by 8.

Total = 48+12 =60
probability= 60/96 = 5/8

5. Good post? |
Agree with supersuj.. right approach

Actually when i say it should be devisible by 4.

then i presume n=4
then (n+1) = 5
(n+2) = 6

so n*(n+1)*(n+2) = 4*5*6 is divisible by 8
but even i was thinking i am missing out multiplication factor of 2*3*4 but supersuj has suggested right approach.

thanks

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