nick1816 wrote:
For how many integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.
A. 9
B. 29
C. 40
D. 49
E. 59
(n-1)*n*(n+1) is a product of 3 consecutive integers.
So, it can be of 2 types
1. even*odd*even
2. Odd*even*odd
Case1: even*odd*even
If one even number is a multiple of 2, the next even number will definitely be a multiple of 4.
Eg: 2*3*4, 10*11*12 etc..
So, number if possible values of n are
3, 5, . . . . 79.
To find number of terms of Arithmetic Progression, use Last term
79 = 3 + (n-1)2
n = 39.
Case2: Odd*even*odd
In this case, the even number SHOULD be a multiple of 8 as the other 2 numbers are odd.
So, favourable values are just the number of multiples of 8 from 2≤n≤80 which are 10 values.
So, total values of n = 39 + 10 = 49
IMO D.
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