AMC 10 Step by Step

2012 AMC 10B Problem 4

P1-10~1 minPrint

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?

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Key insight

Put the two leftover piles together: 4 + 3 = 7 marbles. Seven marbles fill one more bag of 6, leaving 1.

Solution

  1. Ringo's marbles make some full bags of 66, plus 44 extra marbles.
  2. Paul's marbles make some full bags of 66, plus 33 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=74 + 3 = 7 extra marbles.
  4. Seven marbles fill one more bag of 66, with 7−6=17 - 6 = 1 marble left over.

Try it with real numbers: Ringo has 1010 marbles (6+46 + 4) and Paul has 99 (6+36 + 3). Together that is 19=3×6+119 = 3 \times 6 + 1. One marble is left over.

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

Why this works

The leftover of a sum depends only on the leftovers of the parts. Add the leftovers, and if the result is 66 or more, take out another full bag. Mathematicians write this as 4+3=7≡1(mod6)4 + 3 = 7 \equiv 1 \pmod 6, which just means "77 leaves the same leftover as 11 when you divide by 66."

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 44 extra; Paul's fill some whole bags plus 33 extra. When they pool everything, the full bags stay full, and the 4+3=74 + 3 = 7 leftover marbles are combined.

Seven marbles fill one more bag of 66 with 11 marble remaining.

In congruence language: R≡4R \equiv 4 and P≡3(mod6)P \equiv 3 \pmod 6, so R+P≡7≡1(mod6)R + P \equiv 7 \equiv 1 \pmod 6.

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

Review in the AoPS books

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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.