A number with one hundred \(0\)s, one hundred \(1\)s, and one hundred \(2\)s is written on the board. Could this number be a perfect square? You may use the divisibility rules for \(3\) and \(9\): a number is divisible by \(3\) (respectively \(9\)) if the sum of its digits is divisible by \(3\) (respectively \(9\)).
How many divisors does the number \(3^{31}\times 5^{23}\times 7^5\) have?
Recall that \(n! = n\times (n-1)\times (n-2)\times \cdots\times 2\times 1\). Can \(n!\) end in exactly \(5\) zeroes for some \(n\)?