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Q.
Consider $N =7^{11} 13^{29}$. The number of positive integer divisors of $N ^2$ which are less than $N$ but do not divide $N$ is equal to
Permutations and Combinations
Solution:
Let $N = p ^{ r } q ^{ s }$, where $p$ and $q$ are distinct primes.
Then $N ^2= p ^{2 r } q ^{2 s }$, so $N ^2$ has $(2 r +1)(2 s +1)$ divisors. For each divisor less than $N$, there is a corresponding divisor greater than $N$. By excluding the divisor $N$, we see that there must be
$\frac{(2 r +1)(2 s +1)-1}{2}=2 rs + r + s$
divisors of $N ^2$ that are less than $N$. But $N$ has $( r +1)( s +1)$ divisors (including $N$ itself) and because every divisor of $N$ is also a divisor of $N ^2$, so there are
$(2 rs + r + s )-(( r +1)( s +1)-1)= rs$
divisors of $N ^2$, which are less than $N$ but not divisors of $N$.
When $r=11, s=29$, there are $r s=11 \times 29=319$