2013 AMC 10B Problem 25

Only the last two digits of $S$ matter, and those come only from the last two digits of each number Bernardo writes. So we need the last two digits of $N$ in base $5$ and in base $6$.

Names. Let the base-5 form of $N$ end in the digits $d_1 d_0$ and the base-6 form end in $e_1 e_0$. The last digit $d_0$ is the remainder when $N$ is divided by $5$, and $e_0$ is the remainder when $N$ is divided by $6$.

Step 1: the units digit.

1. $N$ is a multiple of $5$ plus $d_0$, so $2N$ is a multiple of $10$ plus $2d_0$. Since $d_0 \le 4$, the units digit of $2N$ is $2d_0$.
2. The units digit of $S$ comes from $d_0 + e_0$, which is at most $4 + 5 = 9$, so there is no carry. We need $d_0 + e_0 = 2d_0$, which means $e_0 = d_0$.
3. So $N$ leaves the same remainder $r$ when divided by $5$ and by $6$, and $r$ is $0, 1, 2, 3$ or $4$. Then $N - r$ is a multiple of both $5$ and $6$, so of $30$:
$$N = 30m + r, \qquad r = 0, 1, 2, 3 \text{ or } 4.$$
(The example $N = 749 = 30 \times 24 + 29$ fails this: its last digits $4$ and $5$ give $9$, but $2N = 1498$ ends in $8$.)

Step 2: the tens digit.

4. Let $k$ be the whole-number part of $N \div 25$. Then $N = 25k + 5d_1 + d_0$, so
$$2N = 50k + 10d_1 + 2d_0.$$
If $k$ is even, $50k$ is a multiple of $100$ and does not touch the last two digits. If $k$ is odd, it adds $50$. So the tens digit of $2N$ is $d_1$ when $k$ is even, and $d_1 + 5$ (dropping any extra $10$) when $k$ is odd.
5. The tens digit of $S$ comes from $d_1 + e_1$. For these to match, we need $e_1 = 0$ when $k$ is even and $e_1 = 5$ when $k$ is odd. (The digit $e_1$ is at most $5$, so no other value can work.)

Step 3: turn this into a condition on $m$.

1. The base-6 tens digit $e_1$ is the remainder when $N \div 6$ (whole-number part) is divided by $6$. Since $N = 30m + r$ with $r < 6$, the whole-number part of $N \div 6$ is $5m$. The remainder of $5m$ divided by $6$:

So $e_1 = 0$ needs $m$ to be a multiple of $6$, and $e_1 = 5$ needs $m$ to be one more than a multiple of $6$.
7. The value of $k$ depends only on $m$: $30m$ is a multiple of $5$, so adding $r \le 4$ never reaches the next multiple of $25$.
8. $N$ has three digits, so $100 \le 30m + r \le 999$, which means $m$ runs from $4$ to $33$. Check the candidates:

The values that work are $m = 12, 24, 30$ and $m = 13, 31$: five values.
9. Each $m$ goes with $5$ choices of $r$, so there are $5 \times 5 = 25$ values of $N$.

Check one: $N = 360$ is $2420$ in base $5$ and $1400$ in base $6$. Then $S = 2420 + 1400 = 3820$, and $2N = 720$. Both end in $20$.

The answer is $\boxed{\textbf{(E)}\ 25}$.

Only the last two digits of each representation affect the last two digits of $S$, so everything comes down to remainders. Handle the units digit first: it cuts the search down to numbers of the form $30m + r$. Then the tens digit becomes a simple even-or-odd check, leaving a short table instead of hundreds of cases.