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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.
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