AMC 8 · 2014 · #21
Grade 6 number-theoryProblem
The 7-digit numbers 74A52B1 and 326AB4C are each multiples of 3. Which of the following could be the value of C?
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: Two $7$-digit numbers, $\underline{7}\underline{4}\underline{A}\underline{5}\underline{2}\underline{B}\underline{1}$ and $\underline{3}\underline{2}\underline{6}\underline{A}\underline{B}\underline{4}\underline{C}$, are each multiples of $3$. The unknown digits $A$, $B$, $C$ are each between $0$ and $9$. Which value among the choices could $C$ be?
Givens: First number: $74A52B1$ is a multiple of $3$; Second number: $326AB4C$ is a multiple of $3$; $A$, $B$, $C$ are single digits ($0$ through $9$); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $8$
Unknowns: The digit $C$ in the second number
Understand
Restated: Two $7$-digit numbers, $\underline{7}\underline{4}\underline{A}\underline{5}\underline{2}\underline{B}\underline{1}$ and $\underline{3}\underline{2}\underline{6}\underline{A}\underline{B}\underline{4}\underline{C}$, are each multiples of $3$. The unknown digits $A$, $B$, $C$ are each between $0$ and $9$. Which value among the choices could $C$ be?
Givens: First number: $74A52B1$ is a multiple of $3$; Second number: $326AB4C$ is a multiple of $3$; $A$, $B$, $C$ are single digits ($0$ through $9$); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $8$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
Two unknowns ($A$, $B$) appear in BOTH numbers, but only $C$ shows up in the second one. Tool #7 (Identify Subproblems) handles this by splitting the question into two clean jobs: (1) extract what the first number's divisibility rule says about $A + B$, then (2) plug that fact into the second number's rule to pin down $C$ alone. Tool #3 (Eliminate Possibilities) is what we use at the end — scan the five answer choices and keep only the one that fits the rule we derived for $C$. Tool #6 (Guess and Check) is the natural fallback: try each answer choice with the divisibility rule and see which one passes.
Execute — Answer: A
4.OA.B.4 Step 1 - Subproblem 1: turn the first number's divisibility into a fact about $A + B$.
- Add the known digits of $74A52B1$: $7 + 4 + 5 + 2 + 1 = 19$.
- So the full digit sum is $19 + A + B$.
- For the number to be a multiple of $3$, this sum must be a multiple of $3$.
- Since $19 = 18 + 1$ and $18$ is already a multiple of $3$, the leftover $1 + A + B$ must be a multiple of $3$.
💡 The Grade 4 "recognize multiples" idea: a sum is a multiple of $3$ exactly when the extra piece on top of a known multiple of $3$ is itself a multiple of $3$.
4.OA.B.4 Step 2 - Subproblem 2: do the same for the second number.
- Add the known digits of $326AB4C$: $3 + 2 + 6 + 4 = 15$.
- The full digit sum is $15 + A + B + C$.
- Since $15$ is already a multiple of $3$, the leftover $A + B + C$ must itself be a multiple of $3$.
💡 Same divisibility rule, same Grade 4 multiple-recognition move.
6.EE.B.5 Step 3 - Combine the two facts.
- Step 1 says $1 + A + B$ is a multiple of $3$, so $A + B$ is one LESS than a multiple of $3$ — in other words, $A + B$ leaves a remainder of $2$ when divided by $3$.
- Step 2 says $A + B + C$ is a multiple of $3$.
- Subtract: $C = (A + B + C) - (A + B)$ is the difference between a multiple of $3$ and a number that leaves remainder $2$, so $C$ itself must leave remainder $1$ when divided by $3$ (because $0 - 2 = -2 \equiv 1$ on a $0,1,2$ cycle).
💡 Subtracting one constraint from another to isolate a single variable is a Grade 6 "use equations as questions about which values work" move — exactly the substitution idea.
4.OA.B.4 Step 4 - Apply the rule "$C$ leaves remainder $1$ when divided by $3$" to the five answer choices.
- Compute each choice mod $3$ and keep the one whose remainder is $1$.
💡 Checking each candidate's remainder against the required remainder is the Grade 4 "factors and multiples" skill in eliminate-possibilities form.
6.EE.B.5 Step 5 Only $C = 1$ leaves remainder $1$ on division by $3$, so that is the answer.
💡 Reading off the lone surviving choice after elimination is the final Grade 6 "which value makes the equation true" step.
4.OA.B.4 Subproblem 1: turn the first number's divisibility into a fact about $A + B$. Ad 4.OA.B.4 Subproblem 2: do the same for the second number. Add the known digits of $326AB4 6.EE.B.5 Combine the two facts. Step 1 says $1 + A + B$ is a multiple of $3$, so $A + B$ 4.OA.B.4 Apply the rule "$C$ leaves remainder $1$ when divided by $3$" to the five answer 6.EE.B.5 Only $C = 1$ leaves remainder $1$ on division by $3$, so that is the answer. Review
Reasonableness: Test the answer with a concrete $A$ and $B$. Pick $A = 1$, $B = 1$, so $A + B = 2$ (leaves remainder $2$ mod $3$, matching Step 1). First number: $7411521$. Digit sum $= 7+4+1+1+5+2+1 = 21 = 3 \times 7$, divisible by $3$. Second number with $C = 1$: $3261141$. Digit sum $= 3+2+6+1+1+4+1 = 18 = 3 \times 6$, divisible by $3$. Both checks pass, so $C = 1$ really works and choice (A) is correct.
Alternative: Tool #6 (Guess and Check) directly on the choices: pick any sample $A,B$ with $A+B \equiv 2 \pmod 3$ — say $A = B = 1$ — and test each choice for $C$ in the second digit sum $15 + 1 + 1 + C = 17 + C$. We need $17 + C$ divisible by $3$, i.e. $C \equiv 1 \pmod 3$. Trying the choices: $17+1=18$ (yes), $17+2=19$ (no), $17+3=20$ (no), $17+5=22$ (no), $17+8=25$ (no). Only (A) $C = 1$ works.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Applying the divisibility-by-$3$ rule (digit sum is a multiple of $3$) to both $7$-digit numbers and checking which answer choices leave remainder $1$ when divided by $3$.)6.EE.B.5Understand solving an equation as a process of answering a question: which values from a specified set make the equation true (Subtracting the first constraint ($A+B \equiv 2 \pmod 3$) from the second ($A+B+C \equiv 0 \pmod 3$) to isolate $C$, and identifying which of the five listed values for $C$ satisfies the resulting condition.)
⭐ This AMC 8 problem only needs the Grade 4 divisibility-by-$3$ trick (add the digits) plus a Grade 6 "which value fits both rules?" check that you already know!
⭐ This AMC 8 problem only needs the Grade 4 divisibility-by-$3$ trick (add the digits) plus a Grade 6 "which value fits both rules?" check that you already know!
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