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AMC 8 · 2014 · #8

Grade 5 number-theory
Archetype Small-Domain Constrained Search
divisibility-rulesdigit-sumplace-value digit-constraintssystematic-enumeration ↑ Prerequisites: divisibility-rulesplace-value
📏 Short solution 💡 2 insights
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Problem

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker \textdollar1‾A‾2‾\textdollar\underline{1} \underline{A} \underline{2}. What is the missing digit AA of this 33-digit number?

(A) 0(B) 1(C) 2(D) 3(E) 4\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4

Pick an answer.

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

#3 Eliminate Possibilities 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 $11$.
$$\overline{1A2} = 11 \times (\text{amount each member paid})$$

💡 Reading a word problem and recognizing "same amount $\times$ number of people = total" is a Grade 4 multi-step word-problem skill.

#3 Eliminate Possibilities 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.
$$\{102,\ 112,\ 122,\ 132,\ 142\}$$

💡 Listing every candidate up front turns an "open" problem into a finite checklist — the core move of Tool #3.

#6 Guess and Check 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 $132$.
$$11 \times 10 = 110,\ \ 11 \times 11 = 121,\ \ 11 \times 12 = 132,\ \ 11 \times 13 = 143$$

💡 Building the $11$ times table step by step is Grade 4 multi-digit multiplication — and it makes the matching value jump out visually.

#6 Guess and Check 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.
$$132 \div 11 = 12 \quad \text{(remainder } 0\text{)}$$

💡 Dividing a $3$-digit number by a $2$-digit number with $0$ remainder is exactly the Grade 5 division standard.

#3 Eliminate Possibilities 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 $\textbf{(D)}$.
$$\overline{1A2} = 132 \;\Rightarrow\; A = 3 \;\Rightarrow\; \textbf{(D)}$$

💡 Once $4$ candidates are eliminated, the remaining one is forced — Tool #3 in its purest form.

[1] #3 4.OA.A.3 Translate the word problem into a math condition. "$11$ members each paid the sa
[2] #3 4.OA.A.3 List the five candidate totals, one for each answer choice. Replacing $A$ with $
[3] #6 4.NBT.B.5 Find the multiples of $11$ near this range so we have something concrete to comp
[4] #6 5.NBT.B.6 Verify by direct division. Divide the surviving total $132$ by $11$ to confirm i
[5] #3 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.3 Solve 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.5 Multiply 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.6 Find 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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