AMC 8 · 2012 · #16
Grade 4 number-theoryProblem
Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
Pick an answer.
AMC 8 2012 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: Use each of the digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once to build two five-digit numbers whose sum is as large as possible. Which one of the answer choices could be one of those two numbers?
Givens: Ten digits available: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$; Each digit must be used exactly once across the two numbers; Both numbers are five-digit numbers (so the ten-thousands digit cannot be $0$); Answer choices: (A) $76531$, (B) $86724$, (C) $87431$, (D) $96240$, (E) $97403$
Unknowns: Which answer choice can appear as one of the two numbers in a maximum-sum arrangement
Understand
Restated: Use each of the digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ exactly once to build two five-digit numbers whose sum is as large as possible. Which one of the answer choices could be one of those two numbers?
Givens: Ten digits available: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$; Each digit must be used exactly once across the two numbers; Both numbers are five-digit numbers (so the ten-thousands digit cannot be $0$); Answer choices: (A) $76531$, (B) $86724$, (C) $87431$, (D) $96240$, (E) $97403$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
The sum of two five-digit numbers is the sum of five place-value columns: ten-thousands, thousands, hundreds, tens, ones. Tool #7 (Identify Subproblems) lets us treat each column as its own mini-question — "which two digits go here?" — because a digit in the ten-thousands column counts $10{,}000$ times while a digit in the ones column counts only $1$ time, so the columns don't trade off against each other. Tool #9 (Easier Problem) sanity-checks this with a smaller version (two 2-digit numbers from digits $0$–$3$). Once we know which digit pair belongs in each column, Tool #3 (Eliminate Possibilities) sweeps through the answer choices and crosses off any number that puts a digit in the wrong column.
Execute — Answer: C
4.NBT.A.2 Step 1 - Try the easier version first.
- Use digits $0, 1, 2, 3$ to make two two-digit numbers with the largest sum.
- Pair the two biggest digits ($3$ and $2$) in the tens column and the two smallest ($1$ and $0$) in the ones column: $31 + 20 = 51$.
- Any other arrangement (like $30 + 21 = 51$ — same digits in each column, just swapped) gives at most $51$, while moving a big digit to the ones column (like $13 + 20 = 33$) is much worse.
- The rule: put the biggest digits in the biggest place values.
💡 A small test case makes the place-value rule obvious before we trust it on the full problem.
4.NBT.A.1 Step 2 - Apply the same rule to the real problem: assign the two largest remaining digits to each column, working from left (largest place value) to right.
- The two biggest digits, $9$ and $8$, must sit in the ten-thousands column.
- Then $7$ and $6$ go in the thousands column, $5$ and $4$ in the hundreds, $3$ and $2$ in the tens, and $1$ and $0$ in the ones.
💡 Each column is its own subproblem: "which two digits maximize this column's contribution?" Answer: the two biggest still available.
4.NBT.B.4 Step 3 - Confirm the maximum sum (a quick reasonableness step).
- Each column contributes (sum of its two digits) $\times$ (place value), and the column sums are forced — $9{+}8 = 17$, $7{+}6 = 13$, $5{+}4 = 9$, $3{+}2 = 5$, $1{+}0 = 1$ — so the maximum sum is fixed regardless of how each pair is split between the two numbers.
💡 Place value tells us the ten-thousands column is worth $10{,}000$ times a ones-column digit, so winning the left columns matters most.
4.NBT.A.2 Step 4 - Eliminate answer choices that put a digit in the wrong column.
- Check each option place by place against the table from Step 2.
💡 With the column rule in hand, each wrong choice is killed by a single mismatched digit — fast multiple-choice elimination.
4.NBT.B.4 Step 5 - Confirm choice (C) actually works by building its partner number from the leftover digits.
- If one number is $87431$, the other must use $9, 6, 5, 2, 0$ — and placing those in the same column rule gives $96520$.
- Check: $87431 + 96520 = 183951$, matching the maximum from Step 3.
💡 A valid answer must come with a valid partner. Building it confirms (C) is realizable, not just rule-compatible.
4.NBT.A.2 Try the easier version first. Use digits $0, 1, 2, 3$ to make two two-digit numb 4.NBT.A.1 Apply the same rule to the real problem: assign the two largest remaining digits 4.NBT.B.4 Confirm the maximum sum (a quick reasonableness step). Each column contributes ( 4.NBT.A.2 Eliminate answer choices that put a digit in the wrong column. Check each option 4.NBT.B.4 Confirm choice (C) actually works by building its partner number from the leftov Review
Reasonableness: The maximum sum $183951$ has six digits, which makes sense: two five-digit numbers near $100{,}000$ each should sum to roughly $200{,}000$. The leading digit of the sum is $1$ because $9 + 8 = 17$ carries a $1$ into a sixth column. Choice (C) $= 87431$ has its biggest digit on the left and smallest on the right, mirroring the place-value rule. The partner $96520$ does the same. Both pass the smell test.
Alternative: Tool #3 (Eliminate Possibilities) on its own, without doing Step 1 first: just by knowing "big digits go on the left", the ten-thousands digit must be $8$ or $9$, instantly killing (A) $76531$. Then for the remaining four choices, scan left to right and stop at the first digit that's smaller than something still available for that column. (B) fails at hundreds ($7$ should have been in the thousands column), (D) fails at hundreds ($2$ instead of $4$ or $5$), (E) fails at tens ($0$ instead of $2$ or $3$). Only (C) survives — no need to compute any sums.
CCSS standards used (min grade 4)
4.NBT.A.1Recognize that in a multi-digit number, a digit in one place represents ten times what it represents in the place to its right (Justifying why the largest digits must go in the highest place values — a ten-thousands digit is worth $10{,}000$ times a ones digit.)4.NBT.A.2Read, write, and compare multi-digit whole numbers using base-ten numerals (Checking each answer choice digit by digit against the required digit pair for its place-value column.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the maximum sum $87431 + 96520 = 183951$ and verifying it matches the column-sum total.)
⭐ This AMC 8 problem only needs the Grade 4 place-value idea you already know: the digit on the far left is worth way more than the one on the right, so put the biggest digits there!
⭐ This AMC 8 problem only needs the Grade 4 place-value idea you already know: the digit on the far left is worth way more than the one on the right, so put the biggest digits there!
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