### Instruction: Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?### Response: This problem involves counting the divisors of a number and comparing them to another number.
I notice that $n^2$ has the same prime factors as $n$, but with twice the exponents.
So, $n^2=2^{62}3^{38}.$
To count the divisors of $n^2$, I can use the formula that says the number of divisors is the product of one more than the exponents of the prime factors.
So, the number of divisors of $n^2$ is $(62+1)(38+1)=2403.$
Half of these divisors are less than $n$ and half are greater than $n$, by symmetry.
So, the number of divisors of $n^2$ that are less than $n$ is $2403/2=1201.5.$
But this is not an integer, so I have to round down to the nearest integer.
So, the number of divisors of $n^2$ that are less than $n$ is $1201.$
But this also includes the divisors of $n$, which I do not want to count.
To count the divisors of $n$, I can use the same formula as before.
The number of divisors of $n$ is $(31+1)(19+1)=640.$
So, the number of divisors of $n^2$ that are less than $n$ but do not divide $n$ is $1201-640=561.$