If n is an integer between 1 and 96 (inclusive), what is the probability that n×(n+1)×(n+2) is divisible by 8?
1/4
1/2
5/8
3/4
7/8
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eshwarjayanth
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MarkMeier20
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When is n*(n+1)*(n+2) divisible by 8?
1. When n is divisible by 8 (12 cases)
2. When n+1 is divisible by 8 (12 cases)
3. When n+2 is divisible by 8 (12 cases)
4. When n is divisible by 4 (24 cases)
Therefore n*(n+1)*(n+2) is divisble by 8 in 60 of 96 cases.
60/97=5/8
1. When n is divisible by 8 (12 cases)
2. When n+1 is divisible by 8 (12 cases)
3. When n+2 is divisible by 8 (12 cases)
4. When n is divisible by 4 (24 cases)
Therefore n*(n+1)*(n+2) is divisble by 8 in 60 of 96 cases.
60/97=5/8
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Frankenstein
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Hi,
n is either even or odd
n is even -> n=2k. So, n(n+1)(n+2) = 2k(2k+1)(2k+2) = 4k(k+1)(2k+1).
k and k+1 are consecutive integers. So, one of them will always be even . Hence k(k+1) is even.
So, whenever n is even, n(n+1)(n+2) is multiple of 8.
Even numbers between 1 and 96 are 96/2 = 48
n is odd -> n is odd, (n+2) is odd. So, for n(n+1)(n+2) to be a multiple of 8, (n+1) should be a multiple of 8.
So, n+1 can be 8,12,..96 -> 12
So, probability is (48+12)/96 = 5/8.
n is either even or odd
n is even -> n=2k. So, n(n+1)(n+2) = 2k(2k+1)(2k+2) = 4k(k+1)(2k+1).
k and k+1 are consecutive integers. So, one of them will always be even . Hence k(k+1) is even.
So, whenever n is even, n(n+1)(n+2) is multiple of 8.
Even numbers between 1 and 96 are 96/2 = 48
n is odd -> n is odd, (n+2) is odd. So, for n(n+1)(n+2) to be a multiple of 8, (n+1) should be a multiple of 8.
So, n+1 can be 8,12,..96 -> 12
So, probability is (48+12)/96 = 5/8.
Cheers!
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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