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2012 AMC 10B Problem 4

2012 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Modular Arithmetic · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

Remainders add: 4 + 3 = 7 marbles left over combine to fill one more bag, leaving 7 - 6 = 1.

Solution

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 $\boxed{\textbf{(A)}\ 1}$ .

Why this works

The remainder of a sum depends only on the remainders of the parts. Add the remainders and reduce once more by the modulus; the full bags never matter. This is the everyday meaning of modular arithmetic.

The trap

Answering 7 (or picking 5 as the largest available choice) by forgetting that 7 leftover marbles fill another bag.

Common mistakes

Answering 7 (or picking 5 as the largest available choice) by forgetting that 7 leftover marbles fill another bag.
Subtracting the remainders ( $4 - 3 = 1$ ), which happens to give the right number here but is the wrong operation.

Techniques

Set up the equation/formula and compute; no special trick needed