AMC 10 · 2003 · #11
Grade 6 arithmeticThe sum of the two 5-digit numbers AMC10 and AMC12 is 123422. What is A+M+C?
Pick an answer.
AMC 10 2003 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 $5$-digit numbers are written as $AMC10$ and $AMC12$, where $A$, $M$, and $C$ are single digits that are the same in both numbers. Their sum is $123422$. Find $A+M+C$.
Givens: $AMC10$ is the $5$-digit number with digits $A$, $M$, $C$, $1$, $0$ (in that order).; $AMC12$ is the $5$-digit number with digits $A$, $M$, $C$, $1$, $2$ (in that order).; $A$, $M$, and $C$ stand for the same digits in both numbers.; $AMC10 + AMC12 = 123422$.; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$.
Unknowns: The value of $A+M+C$.
Understand
Restated: Two $5$-digit numbers are written as $AMC10$ and $AMC12$, where $A$, $M$, and $C$ are single digits that are the same in both numbers. Their sum is $123422$. Find $A+M+C$.
Givens: $AMC10$ is the $5$-digit number with digits $A$, $M$, $C$, $1$, $0$ (in that order).; $AMC12$ is the $5$-digit number with digits $A$, $M$, $C$, $1$, $2$ (in that order).; $A$, $M$, and $C$ stand for the same digits in both numbers.; $AMC10 + AMC12 = 123422$.; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #5 Look for a Pattern, #11 Work Backwards
The two numbers look almost the same, so first use Tool #5 (Look for a Pattern) to spot what actually changes: only the very last digit, $0$ versus $2$. That means the second number is just $2$ bigger than the first, so instead of two mystery numbers there is really only one. Tool #4 (Introduce a Variable) then names that single number $N$, turning the whole puzzle into one clean equation. Finally Tool #11 (Work Backwards) unwinds the equation — undo the $+2$, then undo the doubling — to recover $N$, and the digits of $N$ hand you $A$, $M$, and $C$ directly.
Execute — Answer: E
4.NBT.A.2 Step 1 See that the two numbers are twins
- Compare $AMC10$ and $AMC12$ digit by digit.
- The first four digits are identical — $A$, $M$, $C$, $1$ — and only the last digit differs: $0$ in the first number and $2$ in the second.
- Changing the units digit from $0$ to $2$ adds exactly $2$.
- So the second number is simply the first number plus $2$: $AMC12 = AMC10 + 2$.
💡 If two numbers match in every place except the units, their difference is just the difference of those last digits.
6.EE.B.6 Step 2 Name the number, write one equation
- Let $N$ stand for the first number $AMC10$.
- Then the second number is $N+2$, and the given sum becomes $N + (N+2)$.
- Adding the two copies of $N$ gives $2N$, so the sum is $2N+2$.
- Setting this equal to $123422$ turns the whole problem into a single equation.
💡 Two nearly equal numbers add up to about twice one of them, so naming that one number collapses two unknowns into one.
6.EE.B.7 Step 3 Undo the +2, then undo the doubling
- Work backwards from the sum.
- First remove the extra $2$: $2N = 123422 - 2 = 123420$.
- Then undo the doubling by halving: $N = 123420 \div 2 = 61710$.
- So the first number $AMC10$ equals $61710$.
💡 Reversing each operation that built the sum peels the equation back to the single number hiding inside.
4.NBT.A.2 Step 4 Read off the digits and add
- The number $AMC10$ equals $61710$, so line the digits up with $A$, $M$, $C$, $1$, $0$: $A=6$, $M=1$, $C=7$, and the last two digits are indeed $1$ and $0$, exactly as the pattern requires.
- Therefore $A+M+C = 6+1+7 = 14$, which is choice (E).
💡 Once the number is known, each letter is simply the digit sitting in its place-value slot.
4.NBT.A.2 Compare $AMC10$ and $AMC12$ digit by digit. The first four digits are identical 6.EE.B.6 Let $N$ stand for the first number $AMC10$. Then the second number is $N+2$, and 6.EE.B.7 Work backwards from the sum. First remove the extra $2$: $2N = 123422 - 2 = 1234 4.NBT.A.2 The number $AMC10$ equals $61710$, so line the digits up with $A$, $M$, $C$, $1$ Review
Reasonableness: Check the answer both ways. First, the found digits give $A=6$, $M=1$, $C=7$, so $AMC10 = 61710$ and $AMC12 = 61712$; their sum is $61710 + 61712 = 123422$, matching the problem exactly. Second, the fixed last two digits $1$ and $0$ of $AMC10$ came out right, confirming nothing was mis-aligned. Since each of $A$, $M$, $C$ is a single digit, their sum can be at most $27$, so $14$ is a sensible size and it appears as choice (E).
Alternative: Use place value directly (Tool #13, Convert to Algebra): $AMC10 = 10000A + 1000M + 100C + 10$ and $AMC12 = 10000A + 1000M + 100C + 12$. Adding gives $20000A + 2000M + 200C + 22 = 123422$, so $20000A + 2000M + 200C = 123400$. Dividing every term by $200$ yields $100A + 10M + C = 617$, which reads off as $A=6$, $M=1$, $C=7$, again giving $A+M+C = 14$.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing the two numbers digit by digit to see they differ only in the units place, and reading the digits A, M, C back out of the final value 61710.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting N stand for AMC10 and writing the sum of the two numbers as the expression 2N + 2.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving 2N + 2 = 123422 by removing the 2 and halving to find N = 61710.)
⭐ The two numbers are identical except for the last digit, so one is just $2$ more than the other; that makes the sum equal to twice the first number plus $2$, and undoing those steps gives $AMC10 = 61710$, so $A+M+C = 6+1+7 = 14$.
⭐ The two numbers are identical except for the last digit, so one is just $2$ more than the other; that makes the sum equal to twice the first number plus $2$, and undoing those steps gives $AMC10 = 61710$, so $A+M+C = 6+1+7 = 14$.
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