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by mandeepak » Fri Nov 02, 2007 8:49 pm
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

Is there a shorter way to solve the problem.
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by samirpandeyit62 » Fri Nov 02, 2007 11:17 pm
I will go with D 5/8

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?

total nos of ways in which we can choose n = 96

n(n + 1)(n + 2) will be divisible by 8?

case 1: n = odd then n+2 =odd & n+1 will be even i.e this needs get divided by 8, hence is a multiple of 8 so we have 8..96 = 12 multiples to fill the n+1 pos hence 12 ways

case 2: n is even then n+2 will be even & the product will be divisible by 24 & thus 8

so nos of values that can be used for n= 2....96 (all even nos) i.e 48 nos

total = 48+12 =60 ways

so reqd P =60/96 =5/8
Regards
Samir
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