Q13/021)What is the smallest integer k such that 1^2+2^2+3^2+…+k^2 is divisible by 100?

we have ^2+2^2+3^2+…+k^2 = k(k+1)(2k+1)/6 divisible by 100

so k(k+1)(2k+1) divisible by 600

as k(k+1)(2k+1) is always divisible by 3 so

k(k+1)(2k+1) divisible by 200 = 5^2 * 2^8

now 2k + 1 is odd so k or k+1 should be divisible by 8

that leads to 4 cases
 k  is divisible by 8 and k+1 or 2k+1 by 25
k+ 1 is divisible by 8 and k+1 or 2k+1 by 25

so we look for k mod 8 = 0, k+1 mod 25 = 0 => k = 24

k mod 8 = 0 2k + 1 mod 25 = 0 => k = 112

  k+1 mod 8 = 0 k mod 25 = 0 shall have a larger k = 175
k+1 mod 8 = 0 2k +1 mod 25 = 0 gives k = 87

clearly smallest k = 24