(The competition is meant for 14 to 16 year olds.)
This was from the 2004 exam:
Find the number of digits in N where N is the product of all positive divisors of 100,000,000. (Note: For any positive integer A, we regard A itself as one of the divisors.)
The number A=200420052006...2040 is formed when one puts the consecutive integers from 2004 to 2040 together. What is the remainder when A is divided by 9?
solution discussion for the first one:
Divisors come in pairs -- every divisor x, there must be a divisor n/x, and their product is n. (But if the divisor is a square root of n, then it is the same as the other divisor and it is only counted once.) How many pairs are there?
For n=10 there are two pairs (10-1 5-2), and we note the product is 10^2 = 100.
For n=100 there are five pairs (100-1, 25-4, 20-5, 10-10, 50-2) but one pair is degenerate and the product is 100^4 * 10 (=10^9).
For n=1000, there are eight pairs. Divisor product = 1000^8.
For n=10^8, the script I wrote in Mathematica tells me there are 81 divisors and the divisor product is 325.
The number of divisors for 10^n has an interesting trend, too. It seems to be (n+1)^2.