It has 63 factors. The sum of the divisors should be 51,181. I'm only able to do that quickly using the multiplicative property of the sigma function, which no one would ever be expected to know on the GMAT. I can get there from scratch as follows:

14,400 = 2^6 * 3^2 * 5^2

It has nine odd divisors:
1, 3, 5, 3^2, 5^2, 3*5, 3^2*5, 3*5^2, 3^2*5^2

Let's add these up and call it S:

S = 1+3+5+3^2+5^2+3*5+3^2*5 +3*5^2 + 3^2*5^2
S = 403

Now, we get the rest of the divisors by multiplying each of the odd divisors by 2^1, 2^2, 2^3, 2^4, 2^5 and 2^6. That gives us all 63 divisors. When we multiply the odd divisors by 2^1 and add together the nine new divisors we get, the sum will just be 2^1 * S. When we do the same for 2^2, the sum will be 2^2 * S. And so on. So the sum of all the divisors must be:

S + 2^1*S + 2^2*S + ... + 2^6*S
= (1 + 2 + 4 + 8 + 16 + 32 + 64)*S
= 127*S
= 127*403
= 51,181

That's like a 900-level GMAT question though!

gabriel wrote: Consider a number N which is divided into its prime factors as N=a^x*b*y*c^z . The sum of all the divisors of such a number is given by the formula

(a^0+a^1+a^2+......+a^x)(b^0+b^1+.....+b^y)(c^0+c^1+c^2+....+c^z)

There's a simpler formula which saves you from needing to do those sums. sigma is the name given, in Number Theory, to the function that sums the divisors.

If N = a^x*b*y*c^z, where a, b and c are different primes:
then sigma(N) = [(a^(x+1) - 1)/(a-1)]*[(b^(y+1) - 1)/(b-1)]*[(c^(z+1) - 1)/(c-1)]

So for N = 2^6*3^2*5^2

sigma(N) = [2^7 - 1]*[(3^3 - 1)/2]*[(5^3 - 1)/4] = 127*13*31

Not that you'll ever need to use that on the GMAT.