AMC 8 · 2014 · #8

Grade 5 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
divisibility-rulesdigit-sumplace-value digit-constraintssystematic-enumeration ↑ Prerequisites: divisibility-rulesplace-value
📏 Short solution 💡 2 insights
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Problem
Eleven Math Club members each paid the same whole-dollar amount for a guest speaker. The total they paid is a 3-digit number written as $1A2, where A is an unknown single digit. Find A.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

There are only 5 candidate values for A (0, 1, 2, 3, 4), which gives exactly 5 candidate totals: 102, 112, 122, 132, 142. Tool #3 (Eliminate Possibilities) is the natural first move for any AMC multiple-choice problem with a small finite candidate set — we just test each total for divisibility by 11 and cross out the ones that fail. Tool #6 (Guess and Check) is the engine inside each test: divide the candidate by 11 and see whether the result is a whole number. No algebra and no divisibility rule are needed — direct checking is the fastest, most concrete path for a 5th-grader.

1STEP 1

Because 11 members split the total evenly, 1A2 must be a multiple of 11 — it leaves no remainder.

1A2 = 11 × (amount each member paid)
2STEP 2

Substitute each choice 0, 1, 2, 3, 4 for A to get the five candidate totals 102, 112, 122, 132, 142.

{102, 112, 122, 132, 142}
3STEP 3

Count up the 11 times table — 110, 121, 132, 143 — and the only one inside our candidate set is 132.

11 × 10 = 110, 11 × 11 = 121, 11 × 12 = 132, 11 × 13 = 143
4STEP 4

Divide the survivor to check: 132 ÷ 11 = 12 exactly, so it splits into 11 equal shares.

132 ÷ 11 = 12 (remainder 0)
5STEP 5

The matching total is 132, so its tens digit gives A = 3 — choice (D).

1A2 = 132 → A = 3 → (D)
Answer
3
Each member paid 12and12×11=132—asensibleper−personguest−speakerfeeandthetotalexactlymatches12 and 12 × 11 = 132 — a sensible per-person guest-speaker fee and the total exactly matches1A2 with A = 3. Quickly eliminate the others: 102 ÷ 11 ≈ 9.27, 112 ÷ 11 ≈ 10.18, 122 ÷ 11 ≈ 11.09, 142 ÷ 11 ≈ 12.91 — none whole, so only 132 works.
💡Key takeaway

With just 5 answer choices, you don't need a divisibility rule — try each one and divide. That's Grade 5 division, and AMC 8 rewards it.

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