AMC 10 · 2012 · #4

Easy mode Grade 4

Problem

Ringo puts his marbles into bags of $6$ . He has $4$ marbles left over. Paul puts his marbles into bags of $6$ too. He has $3$ marbles left over. They pour all their marbles together and make as many full bags of $6$ as they can. How many marbles are left over now?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Ringo has some marbles that leave 4 left over when packed 6 per bag. Paul has some marbles that leave 3 left over when packed 6 per bag. They combine all their marbles, pack them 6 per bag again, and we want the number left over.

Givens: Ringo's marbles leave a remainder of 4 when divided into bags of 6.; Paul's marbles leave a remainder of 3 when divided into bags of 6.; The pooled marbles are packed the same way, 6 per bag.

Unknowns: How many marbles are left over after pooling and packing 6 per bag.

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #13 Convert to Algebra

We do not know how many marbles each boy has, only the leftovers. Naming each count with a variable lets us write the 'bags of 6' fact as an equation, add the two counts, and read off the new leftover without ever needing the exact totals.

Execute — Answer: A

#4 Introduce a Variable 4.OA.A.3 Step 1

Write each count with its leftover

Let Ringo have some full bags plus 4 loose marbles, and Paul have some full bags plus 3 loose marbles.
Using a for Ringo's number of full bags and b for Paul's, Ringo has 6a + 4 marbles and Paul has 6b + 3 marbles.

$$R = 6a + 4, \quad P = 6b + 3$$

💡 A remainder just names the loose marbles left after making as many full bags as possible.

#13 Convert to Algebra 4.OA.A.3 Step 2

Add both piles together

Pool the marbles by adding the two counts: (6a + 4) + (6b + 3) = 6a + 6b + 7.
The 6a and 6b are already whole bags, so the only loose marbles come from 4 + 3 = 7.

$$R + P = 6a + 6b + 7$$

💡 Full bags stay full when you combine piles, so only the leftovers can mix into new bags.

#13 Convert to Algebra 4.NBT.B.6 Step 3

Repack the 7 loose marbles

Seven loose marbles is enough for one more full bag with 1 marble to spare, since 7 = 6 + 1.
So the total becomes 6a + 6b + 6 + 1 = 6(a + b + 1) + 1, which is a whole number of bags plus 1 left over.
The answer is (A).

$$6a + 6b + 7 = 6(a + b + 1) + 1$$

💡 The leftover of a sum is just the leftover of adding the two leftovers.

[1] #4 4.OA.A.3 Let Ringo have some full bags plus 4 loose marbles, and Paul have some full bags

[2] #13 4.OA.A.3 Pool the marbles by adding the two counts: (6a + 4) + (6b + 3) = 6a + 6b + 7. Th

[3] #13 4.NBT.B.6 Seven loose marbles is enough for one more full bag with 1 marble to spare, sinc

Review

Reasonableness: Test with the smallest counts: Ringo could have 4 marbles (0 bags, 4 left) and Paul could have 3 marbles (0 bags, 3 left). Together that is 7 marbles, which makes one bag of 6 with 1 left over. That matches answer (A).

Alternative: Skip the variables and reason only with leftovers: the combined leftover is 4 + 3 = 7, and since 7 is more than 6 you pull out one more full bag, leaving 7 - 6 = 1.

CCSS standards used (min grade 4)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Representing each boy's marble count as full bags plus a leftover and combining the two counts.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Finding the remainder when the pooled total is divided into bags of 6.)

⭐ To find the leftover of a sum, just add the leftovers and, if they reach a full bag, take that bag out too.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Write each count with its leftover

Add both piles together

Repack the 7 loose marbles

Review

CCSS standards used (min grade 4)

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