AMC 8 · 2013 · #1

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

multiplesdivisibility-rulesmental-arithmetic systematic-enumeration ↑ Prerequisites: multi-digit-arithmeticmultiples

📏 Short solution 💡 2 insights

Problem

Danica has 23 model cars and wants to arrange them in rows of exactly 6 each, with no empty seats and no leftovers. She needs to buy a few more cars to make this possible. What is the smallest number of additional cars she must buy?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Rows of exactly 6 with no leftovers means the total must be a multiple of 6. Tool #2 (Systematic List) is the natural move: list multiples of 6 in order (6, 12, 18, 24, …) and stop at the first one that is at least 23. Tool #3 (Eliminate Possibilities) is a quick AMC sanity check — add each answer choice to 23 and keep only the totals that are multiples of 6, picking the smallest.

1STEP 1

Rows of exactly 6 with nothing left over means the final total N must be a multiple of 6.

N = 6 × k for some whole number k

2STEP 2

List the multiples of 6 in order and stop at the first one that is at least 23.

6, 12, 18, 24, 30, …

3STEP 3

The smallest multiple of 6 above 23 is 24, so the target is 24 cars; subtract the current 23 to get how many she still needs.

24 - 23 = 1

4STEP 4

Add each choice to 23: only 23 + 1 = 24 lands on a multiple of 6, so the answer is (A).

23 + 1 = 24 = 6 × 4 → (A)

Answer

Sanity check: 23 ÷ 6 = 3 remainder 5. That remainder means Danica is 5 cars into a fourth row of 6, so she needs 6 - 5 = 1 more car to complete that row. That matches the answer (A) and also matches the smallest-multiple-of-6-above-23 calculation, 24 - 23 = 1.

💡Key takeaway

This AMC 8 problem only needs the Grade 4 idea of "multiples" that you already know!

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