1. The sum of squares is equal to n(n+1)(2n+1)/6. Now, there are plenty of numbers n for which this sum is a multiple of 4: take n = 8, n = 16 or basically any multiple of 8 and the sum is consistent with the restriction. This is why 1 is insufficient

2. There are 3 numbers that could be considered: 104, 105 and 106. Again, 2 is insufficient.

Now, but the two together and you get that only 105 fits, so the answer is indeed C

How do you get n(n+1)(2n+1)/6. Is it an equation?

The sum 1^2 + 2^2 + .... + n^2 = n(n+1)(2n+1)/6 - this is a well-known formula and I advise you to remember it - it may come in handy for shortcuts and stuff...

DanaJ

2. There are 3 numbers that could be considered: 104, 105 and 106. Again, 2 is insufficient.

Now, but the two together and you get that only 105 fits, so the answer is indeed C

Shouldn't it be 104, the one that fits??

Dana,

do you have more formulae like this one? It could certainly come in handy...
this question would be pretty difficult to solve in under two minutes without the formula.

Since you asked, here are my formulas for commonly used sums:
I. 1 + 2 + 3 + ... + n = n(n+1)/2

II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

III. 1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2 - this is quite easy to remember, since it's the result from formula I raised to the second power.

avenus: Yes, you are right. Sorry, my mistake.