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)

$hspace{.05in}76531$

(B)

$hspace{.05in}86724$

(C)

$hspace{.05in}87431$

(D)

$hspace{.05in}96240$

(E)

$hspace{.05in}97403$

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

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

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}

3STEP 3

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

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 (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

hspace{.05in}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!

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