AMC 8 · 2014 · #8
Easy mode Grade 5Problem
A math club has 11 members. They each pay the same amount of money to a guest speaker.
When you add up everyone's money, the total is a 3-digit number. The total looks like \textdollar1A2 — the hundreds digit is 1, the tens digit is some unknown digit A, and the ones digit is 2.
Since 11 people each paid the same whole-dollar amount, the total must split evenly into 11 equal parts.
What digit goes in the place of A?
Pick an answer.
AMC 8 2014 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.
Toolkit + CCSS Solution
Understand
Restated: 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 $\$\overline{1A2}$, where $A$ is an unknown single digit. Find $A$.
Givens: Number of members $= 11$; Each member paid the same amount; Total paid $= \$\overline{1A2}$, a $3$-digit number with hundreds digit $1$, tens digit $A$, units digit $2$; Answer choices for $A$: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The single digit $A$ (between $0$ and $9$) that makes the total payment work
Understand
Restated: 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 $\$\overline{1A2}$, where $A$ is an unknown single digit. Find $A$.
Givens: Number of members $= 11$; Each member paid the same amount; Total paid $= \$\overline{1A2}$, a $3$-digit number with hundreds digit $1$, tens digit $A$, units digit $2$; Answer choices for $A$: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #6 Guess and Check
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 $5$th-grader.
Execute — Answer: D
4.OA.A.3 Step 1 - Translate the word problem into a math condition.
- "$11$ members each paid the same amount and together they paid $\$\overline{1A2}$" means $\$\overline{1A2}$ split into $11$ equal shares leaves no remainder.
- So $\overline{1A2}$ must be a multiple of $11$.
💡 Reading a word problem and recognizing "same amount $\times$ number of people = total" is a Grade 4 multi-step word-problem skill.
4.OA.A.3 Step 2 - List the five candidate totals, one for each answer choice.
- Replacing $A$ with $0, 1, 2, 3, 4$ in $\overline{1A2}$ gives the full candidate set.
💡 Listing every candidate up front turns an "open" problem into a finite checklist — the core move of Tool #3.
4.NBT.B.5 Step 3 - Find the multiples of $11$ near this range so we have something concrete to compare against.
- Counting up by $11$: $11, 22, 33, \ldots, 99, 110, 121, 132, 143$.
- The only one of these inside our candidate set $\{102, 112, 122, 132, 142\}$ is $132$.
💡 Building the $11$ times table step by step is Grade 4 multi-digit multiplication — and it makes the matching value jump out visually.
5.NBT.B.6 Step 4 - Verify by direct division.
- Divide the surviving total $132$ by $11$ to confirm it splits evenly into $11$ equal shares.
💡 Dividing a $3$-digit number by a $2$-digit number with $0$ remainder is exactly the Grade 5 division standard.
4.OA.A.3 Step 5 - Read off the digit.
- The matching total is $132$, so the tens digit is $A = 3$.
- That is choice $\textbf{(D)}$.
💡 Once $4$ candidates are eliminated, the remaining one is forced — Tool #3 in its purest form.
4.OA.A.3 Translate the word problem into a math condition. "$11$ members each paid the sa 4.OA.A.3 List the five candidate totals, one for each answer choice. Replacing $A$ with $ 4.NBT.B.5 Find the multiples of $11$ near this range so we have something concrete to comp 5.NBT.B.6 Verify by direct division. Divide the surviving total $132$ by $11$ to confirm i 4.OA.A.3 Read off the digit. The matching total is $132$, so the tens digit is $A = 3$. T Review
Reasonableness: Each member paid $\$12$ and $12 \times 11 = 132$ — a sensible per-person guest-speaker fee and the total exactly matches $\$\overline{1A2}$ with $A = 3$. Quickly eliminate the others: $102 \div 11 \approx 9.27$, $112 \div 11 \approx 10.18$, $122 \div 11 \approx 11.09$, $142 \div 11 \approx 12.91$ — none whole, so only $132$ works.
Alternative: Tool #5 (Look for a Pattern) on the units digit: multiples of $11$ in this range are $110, 121, 132, 143$, and their units digits cycle $0, 1, 2, 3$. We need a units digit of $2$, which forces the multiple to be $132$ and so $A = 3$. A more advanced learner could also use the divisibility-by-$11$ rule (alternating digit sum): $1 - A + 2 = 3 - A$ must be a multiple of $11$, giving $A = 3$ as the only single-digit solution.
CCSS standards used (min grade 5)
4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Translating "$11$ members each paid the same amount, total $\$\overline{1A2}$" into the condition "$\overline{1A2}$ must be a multiple of $11$", and listing all five candidate totals.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers (Building the $11$ times table ($11 \times 10 = 110$, $11 \times 11 = 121$, $11 \times 12 = 132$, $11 \times 13 = 143$) to locate the multiple of $11$ inside the candidate range.)5.NBT.B.6Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors (Confirming $132 \div 11 = 12$ with remainder $0$, so the $11$ members each paid a whole number of dollars.)
⭐ 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.
⭐ 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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