Question

Let $S_m$ denote the sum of the squares of the first m natural numbers. For how many values of m < 100, is $S_m$ a multiple of 4?

• A. 50
• B. 25
• C. 36
• D. 24

Answer 1

The correct option is D 24

We need the expression $\frac{\left(n\right(n+1\left)\right(2n+1\left)\right)}{6}$ to be a multiple of 4. For this to occur, the numerator of the above expression should be a multiple of 8. In the expression n (n + 1) (2n + 1), 2n + 1 would always be an odd number. Also, amongst n and (n + 1), one number would be odd and the other would be even. Since, we need n (n + 1) (2n + 1) to be a multiple of 8, we would need either n or (n + 1) to be a multiple of 8(while at the same time it should be below 100). Thus, we get the number series n = 7, 8, 15, 16, 23, 24, .... 95, 96. Since there are 12 multiples of 8 below 100, the required answer is $12\times 2=24$