Problem - 2334
Let $a$, $b$, $c$, $d$, and $e$ be five positive integers. If $ab+c=3115$, $c^2+d^2=e^2$, both $a$ and $c$ are prime numbers, $b$ is even and has $11$ divisors. Find these five numbers
Because $11$ is a prime number, $b$ must be in the form of $k^{10}$ in order to have $11$ divisors. Consider it is even, we can be certain that $b=2^{10}=1024$.
Next, because both $a$ and $c$ are positive prime, $a$ can only be $2$ or $3$. It is easy to show that $2$ does not work, hence $a=3$ and $c=43$. $(c, d, e)$ needs to be a Pythagorean triplet. By the Pythagorean triplet formula (or try and guess), we can determine $d=924$ and $e=925$.
Therefore, in conclusion $(a, b, c, d, e) = \boxed{(3, 1024, 43, 924, 925)}$.