Question

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

• A. 319
• B. 349
• C. 339
• D. 329

Answer 1

Answer: (D)

Let $N=p^r{q}^s$, where $p$ and $q$ are distinct primes.
Then $N^2=p^{2r}{q}^{2s}$, so $N^2$ has $(2r+1)(2s+1)$ 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
$\frac{(2r+1)(2s+1)-1}{2}=2rs+r+s$
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
$(2rs+r+s)-((r+1)(s+1)-1)=rs$
divisors of $N^2$, which are less than $N$ but not divisors of $N$.
When $r=11,s=29$, there are $rs=11\times 29=319$