AMC 10 · 2015 · #3

Grade 4 algebra
linear-equations-two-vardivisibility-rules convert-to-algebracasework ↑ Prerequisites: linear-equations-two-var
📏 Short solution 💡 2 insights
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Problem
One number appears twice and another three times, adding to a known total. Find the unknown number.

Pick an answer.

(A)
8
(B)
11
(C)
14
(D)
15
(E)
18
How to solve
Strategy Eliminate Possibilities

We are not told whether 28 is the number repeated twice or the number repeated three times, so there are only two possible setups. Tool #3 (Eliminate Possibilities) tests each setup and throws out the one that cannot produce a whole number, leaving exactly one valid answer. Tool #4 (Introduce a Variable) names the unknown integer x so we can track what each case forces it to be. The decisive clue is that the other integer must come out whole — only one of the two cases survives that test.

1STEP 1

Name the unknown, list the two cases

There are two possible cases.

Case A: 3 · 28 + 2x = 100, Case B: 2 · 28 + 3x = 100
2STEP 2

Test Case A: 28 three times

The first gives a clean whole 8.

3 · 28 = 84, 100 - 84 = 16, x = 16 ÷ 2 = 8
3STEP 3

Test Case B, eliminate it, read the answer

The second fails to give a whole number, choice (A).

2 · 28 = 56, 100 - 56 = 44, 44 ÷ 3 is not a whole number → (A) 8
Answer
8
Check the winning case directly: the five numbers are 28, 28, 28, 8, 8, and 28+28+28+8+8 = 84 + 16 = 100. That matches the given total, one of the numbers is 28 as required, and both repeated values are integers. The answer 8 is also one of the listed choices.
💡Key takeaway

When you don't know which slot a number fills, try every option and keep only the one that comes out as a whole number.

  • Name the unknown, list the two cases
  • Test Case A: 28 three times
  • Test Case B, eliminate it, read the answer