2012 AMC 10B Problem 4

1. Ringo's marbles make some full bags of $6$, plus $4$ extra marbles.
2. Paul's marbles make some full bags of $6$, plus $3$ extra marbles.
3. When they pool everything, the full bags are still full, so they don't matter. Only the extras need new bags: $4 + 3 = 7$ extra marbles.
4. Seven marbles fill one more bag of $6$, with $7 - 6 = 1$ marble left over.

Try it with real numbers: Ringo has $10$ marbles ($6 + 4$) and Paul has $9$ ($6 + 3$). Together that is $19 = 3 \times 6 + 1$. One marble is left over.

The answer is $\boxed{\textbf{(A)}\ 1}$.

The leftover of a sum depends only on the leftovers of the parts. Add the leftovers, and if the result is $6$ or more, take out another full bag. Mathematicians write this as $4 + 3 = 7 \equiv 1 \pmod 6$, which just means "$7$ leaves the same leftover as $1$ when you divide by $6$."

Ringo's marbles fill some whole bags plus $4$ extra; Paul's fill some whole bags plus $3$ extra. When they pool everything, the full bags stay full, and the $4 + 3 = 7$ leftover marbles are combined.

Seven marbles fill one more bag of $6$ with $1$ marble remaining.

In congruence language: $R \equiv 4$ and $P \equiv 3 \pmod 6$, so $R + P \equiv 7 \equiv 1 \pmod 6$.

The answer is $\boxed{\textbf{(A)}\ 1}$.