If an integer n is to be chosen at random from the integers 1 to 94, inclusive, what is the ...
If an integer n is to be chosen at random from the integers 1 to 94, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
Answer/Solution
59/94
Steps/Work
n(n+1)(n+2) will be divisible by 8 when n is a multiple of 2 or when (n+1) is a multiple of 8.
Thus when n is even, this whole expression will be divisible by 8.
from 1 to 96, there are 47 even integers.
Now when (n+1) is multiple by 8, we have 12 such values for (n+1)
probability that n(n+1)(n+2) will be divisible by 8
= (47 + 12)/94
= 59/94
= 5/8
Ans is E
Thus when n is even, this whole expression will be divisible by 8.
from 1 to 96, there are 47 even integers.
Now when (n+1) is multiple by 8, we have 12 such values for (n+1)
probability that n(n+1)(n+2) will be divisible by 8
= (47 + 12)/94
= 59/94
= 5/8
Ans is E