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
I get 1/2. Can some comment on how to tackle this question. The right answer is D. I don't see it. I figured out from 0-9 which value using the formula above which are divisible by 8. I discoved that 2, 4, 6, 7, 8 are divisible by 8. If 2 is divisible by 8 the same should go for 12, 22, 32 and so on. Counting them up I got 48 numbers divisible by 8. 48/96 = 1/2. Where did I go wrong?
Kaunteya
Probabilty and Divisor
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D
take a look at integer sequence:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
to catch 8*n we have to start with 8n-2, 8n-1, 8n
1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19
Therefore, we have 3 combinations for each 8 consecutive integers.
to catch 8*n we also have to start with other even integers 8n-6, 8n-4
1,[2,3,4],5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
1,2,3,[4,5,6],7,8,9,10,11,12,13,14,15,16,17,18,19
p=3/8+ 2/8=5/8
take a look at integer sequence:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
to catch 8*n we have to start with 8n-2, 8n-1, 8n
1,2,3,4,5,[6,7,8],9,10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,[7,8,9],10,11,12,13,14,15,16,17,18,19
1,2,3,4,5,6,7,[8,9,10],11,12,13,14,15,16,17,18,19
Therefore, we have 3 combinations for each 8 consecutive integers.
to catch 8*n we also have to start with other even integers 8n-6, 8n-4
1,[2,3,4],5,6,7,8,9,10,11,12,13,14,15,16,17,18,19
1,2,3,[4,5,6],7,8,9,10,11,12,13,14,15,16,17,18,19
p=3/8+ 2/8=5/8