Question

Define a number K such that it is the sum of the squares of the first M natural numbers. (i.e. $K=1^2+2^2+\dots \dots +M^2)$ where M < 55. How many values of M exist such that K is divisible by 4?

• A. 10
• B. 11
• C. 12
• D. None of these

Answer 1

The correct option is C 12

The sum of squares of the first n natural numbers is given by $\frac{n(n+1)(2n+1)}{6}$.
For this number to be divisible by 4, the product of n(n + 1) (2n + 1) should be a multiple of 8. Out of n, (n + 1) and (2n + 1) only one of n or (n + 1) can be even and (2n + 1) would always be odd.
Thus. either n or (n + 1) should be a multiple of 8. This happens if we use n = 7, 8, 15, 16, 23, 24, 31, 32, 39, 40, 47, 48. Hence, 12 such numbers.