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 = 96
n(n+1)(n+2) => consecutive integer.. thus mutiples of 6
if n = even.. then n+2 is even.. =>n(n+1)(n+2) = multiples of 8
the number of even integers in given interval = 96/2 = 48
if n+1 = multiples of 8 also satisfy the condition.
the number of multiples of 8 in given interval = 96/8 = 12
Thus, (48+12)/96 = 5/8
D it is.
Just add 2 more cents..
If u not sure why even consecutive intergers are multiples of 8..
An even integer = 2m
n*(n+2) = 2m(2m+2) = 4m(m+1) : either m or m+1 is even. Thus, always multiples of 8.
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