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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(
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11 Nov 2022, 05:08
11 Nov 2022, 05:08
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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
A. 25%
B. 50%
C. 62.5%
D. 72.5%
E. 75%
A. 25%
B. 50%
C. 62.5%
D. 72.5%
E. 75%
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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(
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14 Nov 2022, 07:42
14 Nov 2022, 07:42
1
Expert Reply
n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers
also cases in which (n+1) is a multiple of 8 will be divisible by 8
for ex: n=7,15,23.. i.e. total 12 numbers
so total number of such cases is 48+12=60
probability = 60/96 = 5/8 = 62.5%
Answer: C
also cases in which (n+1) is a multiple of 8 will be divisible by 8
for ex: n=7,15,23.. i.e. total 12 numbers
so total number of such cases is 48+12=60
probability = 60/96 = 5/8 = 62.5%
Answer: C
gmatclubot
If n is an integer from 1 to 96, what is the probability for n*(n+1)*( [#permalink]
14 Nov 2022, 07:42
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