If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?
A.1/4
B.1/2
C.5/8
D.7/8
E.3/4
n(n+1)(n+2) = the product of 3 consecutive integers.
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 a multiple of 8 = 5/8.
The correct answer is
C.
Alternate approach:
Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.
Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.
Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.
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