2012 AMC 10B Problem 4
When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?
Show solution without answering
Key insight
Put the two leftover piles together: 4 + 3 = 7 marbles. Seven marbles fill one more bag of 6, leaving 1.
Solution
- Ringo's marbles make some full bags of , plus extra marbles.
- Paul's marbles make some full bags of , plus extra marbles.
- When they pool everything, the full bags are still full, so they don't matter. Only the extras need new bags: extra marbles.
- Seven marbles fill one more bag of , with marble left over.
Try it with real numbers: Ringo has marbles () and Paul has (). Together that is . One marble is left over.
The answer is .
Why this works
The leftover of a sum depends only on the leftovers of the parts. Add the leftovers, and if the result is or more, take out another full bag. Mathematicians write this as , which just means " leaves the same leftover as when you divide by ."
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Ringo's marbles fill some whole bags plus extra; Paul's fill some whole bags plus extra. When they pool everything, the full bags stay full, and the leftover marbles are combined.
Seven marbles fill one more bag of with marble remaining.
In congruence language: and , so .
The answer is .
Review in the AoPS books
- Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
- Introduction to Number Theory, Chapter 10: Units Digits
- Introduction to Number Theory, Chapter 14: Linear Congruences (§14.4)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.