The sum of the base-$10$ logarithms of the divisors of $10^n$ is $792$. What is $n$?
This is Problem 23 form the 2008 AMC 12B.
My solution: Sum of logarithms is logarithm of all of the arguments, so this sum is just log base 10 of the product of the divisors of 10^n
By some elementary number theory, the product of divisors of 10^n is 10^((n+1)^2/2).
(By someone's suggestion I'll detail what I meant) For any number like 10, you can list out its divisors in pairs, lets say the product of its divisors is P P = 1 * 2 * 5 * 10 P = 10 * 5 * 2 * 1 P^2 = 10 * 10 * 10 * 10 P = 100
notice how P^2 is 10^(number of divisors of 10), so P is 10^(num_divisors)/2
To find the number of divisors: consider the prime factorization: 2^a * 5^b
a can be 0 or 1, same for b. So there are 2 * 2=4 divisors.
I generalized this for 10^n, where the prime factorization is 2^n * 5^n. So there are (n+1)^2 divisors, and therefore 10^(n+1)/2 is their product.
Since we are taking the log base 10 of this, we just get (n+1)^2/2=792
Except in the actual solution, they multiply by n for some reason, but I don't understand why?
My solutions follows solution 2: https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_23
Edit: Ok so I figured out where I messed up with my solution, although the one by Pi Quark is perfectly valid:
I calculated the number of divisors of 10^n, but then raised only 10 to that power, I need to raise 10^n to (n+1)^2/2
This way, because of exponent power rule (multiply the exponents), the exponent of 10 is n(n+1)^2/2, when taking the log becomes just n(n+1)^2/2
Let's just focus on the powers of 5. For a constant power of 2(say, $2^1=2$), we have the choices of divisors: $2,2*5, 2*5^2,\cdots, 2*5^n$. So, the total power of 5 in this product will be $\dfrac{n(n+1)}{2}$. Now since, the total product is $\prod_{a=0}^{n} 2^a*5^{\frac{n(n+1)}{2}} $, the total power of 5 is $(n+1)\dfrac{n(n+1)}{2}$. Note that 2 should have same power as 5 in the product.