## Online Course Discussion Forum

### Week 1 HW AMC 12 Prep Course

Hi, I'm currently stuck on problem 9 on our homework, ie

Bernardo chooses a three-digit positive integer NN and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer SS. For example, if N=749N=749, Bernardo writes the numbers 10,44410,444 and 3,2453,245, and LeRoy obtains the sum S=13,689S=13,689. For how many choices of NN are the two rightmost digits of SS, in order, the same as those of 2N2N?

So far, I've found that N <= 4 mod 30 by looking at the ones digit for  N in base 6, 5, and 10, and solving some congruences.

However, I have no idea where to go from here or how to find out what the possible tens digits of N in base 6 or N in base 5 could be.

I would really appreciate some help!

This question is 2013 AMC 10B #25.

You found that $N$ has remainder $0,1,2,3,4$ in mod $30$.  This is a big step!  If a number $N\equiv 0\pmod{30}$ works, then $N+1$, $N+2$, $N+3$, and $N+4$ all work.  So we only need to check the possibilities where $N\equiv 0\pmod{30}$.  In this case the last digit of $N$ (in base $10$) must be $0$.  So we need to list and check the $10$ cases where $N$ ends with $00$, $10$, $20$, $30$, $\ldots$, $90$.  Checking is pretty easy, and some of these cases have solutions, and some don't.  The final answer is $5$ times the number of these cases that do have solutions.

I see, thanks for the help! Just making sure, we multiply by 5 because of the 5 cases where N is congruent to 0, 1, 2, 3, and 4 mod 30 right?

Exactly.

Actually, sorry, but I'm a bit confused about what you mean by checking. Could you hint me in the right direction or give me an example?