Competition · AMC preparation · step 4 of 4

AMC 8 · 2012 · #16

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

place-valuedigit-constraintssystematic-enumeration caseworkdigit-constraintssystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticplace-value

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

76531

(B)

86724

(C)

87431

(D)

96240

(E)

97403

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Try a smaller version first

Test a smaller case: from 0–3, 31 + 20 = 51 beats 13 + 20 = 33. The rule: put the biggest digits in the biggest place values.

31 + 20 = 51 vs. 13 + 20 = 33

2STEP 2

Assign digits by place value

Same rule, left to right: {9,8} in the ten-thousands, {7,6} thousands, {5,4} hundreds, {3,2} tens, {1,0} ones.

Place & Digit pair ; ten-thousands & {9, 8} ; thousands & {7, 6} ; hundreds & {5, 4} ; tens & {3, 2} ; ones & {1, 0}

Each column is its own subproblem: "which two digits maximize this column's contribution?" Answer: the two biggest still available.

To make the sum of the two five-digit numbers as large as possible, the largest digits must go in the highest place-value columns: 9 and 8 in the ten-thousands column, 7 and 6 in the thousands, 5 and 4 in the hundreds, 3 and 2 in the tens, and 1 and 0 in the ones.

▸ Why?

The sum of the two numbers is a total of five place-value columns, and a digit placed in a higher column counts toward that total far more than the same digit in a lower column, so the sum is largest when the biggest digits fill the highest places.

▸ Why?

When you add the two numbers, you can total each place-value column on its own and then combine them, and each column's contribution is its two digits counted in that column's place value.

▸ Why?

A written number is just each digit counted in its place — eight ten-thousands, seven thousands, and so on — so its value comes entirely from which place each digit sits in.

Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

▸ Why?

The full sum can be found by adding the columns separately and joining the results, because a whole equals the sum of its parts no matter how you group them.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Each column to the left is worth ten times the column on its right, so the ten-thousands place multiplies its digits by ten thousand while the ones place multiplies by only one.

Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

▸ Why?

If a smaller digit ever sat in a higher place while a larger digit sat in a lower place, swapping the two would raise the total, so the largest possible sum keeps the biggest digits in the highest places.

▸ Why?

Swapping those two digits changes the total by (the larger digit minus the smaller) times (the higher place value minus the lower), and both differences are positive, so the swap can only make the total bigger.

Distributive propertya groups of (b and c) is a groups of b plus a groups of c — you can split the group.

4.NBT.A.1Identify Subproblems

3STEP 3

Confirm the maximum sum

Column sums are forced: 9+8=17, 7+6=13, 5+4=9, 3+2=5, 1+0=1, so the maximum sum 183951 is fixed no matter how each pair splits.

17 · 10000 + 13 · 1000 + 9 · 100 + 5 · 10 + 1 = 183951

4STEP 4

Eliminate the misplaced choices

Check each choice place by place against the Step 2 table; cross off any digit in the wrong column. Only (C) survives every column.

(A) 76531 & ten-thousands = 7, not in {9,8} X ; (B) 86724 & hundreds = 7, not in {5,4} X ; (C) 87431 & 8 ∈ {9,8}, 7 ∈ {7,6}, 4 ∈ {5,4}, 3 ∈ {3,2}, 1 ∈ {1,0} ✓ ; (D) 96240 & hundreds = 2, not in {5,4} X ; (E) 97403 & tens = 0, not in {3,2} X

5STEP 5

Build the partner number

Build (C)'s partner from the leftovers 9, 6, 5, 2, 0 → 96520, and 87431 + 96520 = 183951 matches the Step 3 maximum.

87431 + 96520 = 183951 → (C)

Answer

87431

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.

💡Key takeaway

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!

Try a smaller version first
Assign digits by place value
Confirm the maximum sum
Eliminate the misplaced choices
Build the partner number

A parent dashboard for the family lives at sensimlab.com.

Try a smaller version first

Assign digits by place value

Confirm the maximum sum

Eliminate the misplaced choices

Build the partner number

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