# Mock AIME II 2012 Problems/Problem 14

## Problem

Call a number a if for all prime divisors of the , , divides and is not prime. Find the sum of all two digit .

## Solution

Remark that all numbers of the form for are . The proof of this is noting that , which is proven by noting that , and therefore we get . Therefore, some two digit that we get are .

Now, we look for of the form . Note that this means that , which is impossible, so we can’t have and together. Also, this implies that and , so we can’t have times any other prime to give us a . We now look at numbers of the form . We can’t have and together because this would imply that divides , but , so this is impossible. Hence, we look for numbers of the form and . The first case gives us or . Therefore , which gives us . However, gives us no values for , therefore we must have and this gives us to give another :

Now, looking at the form gives us or or . Therefore , but this gives us no possibilities for that are greater than (which don’t fall into the first case), therefore there are none of this form.

Lastly, look at the form where is a one digit prime and is a two digit prime. Start out by looking at . For reasons explained before, this cannot be a if is a two digit prime number. Now, look at numbers of the form . Since , we are going to need for . Therefore . gives which isn’t a two digit prime number, gives which gives a largest value of as . gives us , therefore or which aren’t prime. gives us no two digit , since for is greater than .

Now, look at the form to note that either or . The first case gives us and the second gives us , which implies that which isn’t a two digit prime number. Therefore, we look at the first case. We need for or . We also need or . However, this gives us , which gives us no two digit numbers that work.

Lastly, look at numbers of the form . Since implies that which isn’t a two digit prime number, we must have , and therefore we get numbers of the form . Note that therefore or . The first one gives and both false, and the second one gives and , which the second one isn’t true.

Therefore, we have found our to be which gives us a sum of .