# 2007 iTest Problems/Problem 43

## Problem

Bored of working on her computational linguistics thesis, Erin enters some three-digit integers into a spreadsheet, then manipulates the cells a bit until her spreadsheet calculates each of the following one hundred $9$-digit integers: \begin{align*}700\cdot 712\,\cdot\, &718+320,\\701\cdot 713\,\cdot\, &719+320,\\ 702\cdot 714\,\cdot\, &720+320,\\&\vdots\\798\cdot 810\,\cdot\, &816+320,\\799\cdot 811\,\cdot\, &817+320.\end{align*}

She notes that two of them have exactly $8$ positive divisors each. Find the common prime divisor of those two integers.

## Solution

Notice that each number can be written in form $n(n+12)(n+18) + 320$, where $n$ is an integer from $700$ to $799$. Expanding the polynomial results in $n^3 + 30n^2 + 216n + 320$, and factoring that results in $(n+2)(n+8)(n+20)$.

If a number has eight positive divisors, then it’s prime factorization is in the form $a^7$, $a^3b$, or $abc$, where $a$, $b$, and $c$ are primes. The most likely option is three different prime numbers being multiplied together.

After finding the prime numbers from $702$ to $819$, the values of $n$ that work are $731$ and $749$ (after finding one value, note that both share a factor to find the other value). Thus, the prime factorizations of the two numbers are $733 \cdot 739 \cdot 751$ and $751 \cdot 757 \cdot 769$, so the common prime divisor of the two is $\boxed{751}$.