If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)
A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%
Let f(n) = n*(n+1)*(n+2)
The product of the three consecutive numbers will be divisible by 8 if the first of the 3 consecutive numbers is even as it prime factors out to at least three 2’s.
f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even.
f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even
f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd
However, there is one more case where the function will be divisible by 8 even when the first number of the three consecutive number is odd;
f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8.
f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.
Probability = desired/total
How many functions are there;
f(1) = 1*2*3
f(2) = 2*3*4
f(3) = 3*4*5
f(4) = 4*5*6
…
f(95) = 95*96*97
f(96) = 96*97*98
Total = 96
To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.
Count of even numbers = (96–2)/2+1=47+1=48
Total numbers divisible by 8 = (96–8)/8+1 = 12
Probability=48+1296=6096=0.625=62.5Probability=48+1296=6096=0.625=62.5
C
Jeremiah LaBrash is a programmer and CIO for a CCaaS telecom company based in New York, NY. If you have math or verbal questions you’re having difficulty with and would like Jeremiah LaBrash to solve them and parse them into understandable parts, please leave a comment below or mail jr@thelevel11.com
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