If n is an integer from 1 to 96 (inclusive), how m

Question

If n is an integer from 1 to 96 (inclusive), how many numbers of the form n*(n+1)*(n+2) are divisible by 8?

A. 20
B. 30
C. 40
D. 50
E. 60

Please help me with this one!

Hatakekakashi · Accepted Answer

n is an integer 1 to n inclusive

let n = 2

2x3x4 divisble by 8

n = 4

4x5x6

n=6 
6x7x8

for all even n(n+1)(n+2) is divisible by 
so from 1 to 96 even = 96/2 = 48

now check for odd values

if n = 7 
7x8x9

n=15 
15x16x17

n=23

23x24x25

for n=7,15,23 i.e (k-1) where k = multiple of 8 n(n+1)(n+2) is divisible by 8 from 1 to 86 there are 12 values of k (96/8=12)

thus 48 even + 12 odd = 60

(E) imo

Bunuel · Answer

Check here: https://gmatclub.com/forum/if-an-intege ... 26654.html

rahul16singh28 · Answer

Hi  gmatexam439

First -  n*(n+1)*(n+2)  will be divisible by 8 when n = 2,4,6,8..........96. So, here we have 48 Numbers. So, only Option D & E left.
Now, check out all the numbers when added with 1 should be divisible by 8. (You just have to find 3 Numbers to rule out Option D as we already know from above we have 48 numbers).

N = 7,15,23  etc.

Only option E is there.

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If n is an integer from 1 to 96 (inclusive), how m

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If n is an integer from 1 to 96 (inclusive), how m

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