Q2014/001)Given that 1^2+2^2+3^2+.....+k^2=k*(k+1)*(2k+1)/6,what is the is the smallest integer k such that 1^2+2^2+3^2+.....+k^2 is divisible by 100.

k*(k+1)*(2k+1)/6 must be divisible by 100

or k*(k+1)*(2k+1) must be divisible by 600

600 = 25 * 8 * 3

One number is always divisible by 3 so we need not check for it.  only one that is either k or k + 1 is by 2 ( hence 8) and one by 5 ( hence 25)

So 2k + 1 > = 25 or k > = 12

If k is even then multiple of 8: so candidates 16,24

If k is odd then  multiple of 8 – 1 that is 15, 23

So we need to try 15,16,23,24 so on and checking one by one we get  24 as answer