Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?

We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$. There are $32\times20-1 = 639$ factors of $n$, which clearly are less than $n$, but are still factors of $n$. Therefore, using complementary counting, there are $1228-639=\boxed{589}$ factors of $n^2$ that do not divide $n$.
589
Level 5
Number Theory
Problem: Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n$ but do not divide $n$?

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of positive integer divisors of \( n^2 \) that are less than \( n \) but do not divide \( n \), we start by determining the prime factorization of \( n \). Given \( n = 2^{31}3^{19} \), we can express \( n^2 \) as:

\[
n^2 = (2^{31}3^{19})^2 = 2^{62}3^{38}.
\]

Next, we calculate the total number of positive divisors of \( n^2 \) using the formula for the number of divisors based on prime factorization. If \( n^2 = p_1^{e_1} p_2^{e_2} \), the number of divisors \( d(n^2) \) is given by:

\[
d(n^2) = (e_1 + 1)(e_2 + 1).
\]

For \( n^2 = 2^{62}3^{38} \):

\[
d(n^2) = (62 + 1)(38 + 1) = 63 \times 39 = 2457.
\]

Next, we need to find the number of divisors of \( n^2 \) that are less than \( n \). Since the divisors of \( n^2 \) can be paired such that one divisor in each pair is less than \( n \) and the other is greater than \( n \), exactly half of the divisors of \( n^2 \) will be less than \( n \). Therefore, the number of divisors of \( n^2 \) that are less than \( n \) is:

\[
\frac{d(n^2)}{2} = \frac{2457}{2} = 1228.5.
\]

Since the number of divisors must be an integer, we take the floor of this value, which gives us \( 1228 \) divisors less than \( n \).

Next, we need to find the number of divisors of \( n \). The number of positive divisors of \( n \) is calculated similarly:

\[
d(n) = (31 + 1)(19 + 1) = 32 \times 20 = 640.
\]

Now, we need to identify how many of these divisors of \( n \) are also less than \( n \). Since \( n \) itself is not included in the count of divisors less than \( n \), the number of divisors of \( n \) that are less than \( n \) is:

\[
d(n) - 1 = 640 - 1 = 639.
\]

Finally, we find the number of positive integer divisors of \( n^2 \) that are less than \( n \) but do not divide \( n \) by subtracting the number of divisors of \( n \) that are less than \( n \) from the number of divisors of \( n^2 \) that are less than \( n \):

\[
1228 - 639 = 589.
\]

Thus, the final answer is \( 589 \).<|END_OF_LATENT|> The final answer is $589$.