dvalenz wrote: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?
Thanks
n(n+1)(n+2) = the product of 3 consecutive integers.
For many test-takers, the most efficient approach will be to WRITE IT OUT and LOOK FOR A PATTERN.
1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10
9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18
Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.
Thus, the probability that n(n+1)(n+2) will be multiple of 8 = 5/8.
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https://www.beatthegmat.com/probability-t116280.html
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