AMC 10 · 2016 · #6

Grade 4 arithmetic
place-valuedigit-sumdigit-constraints extremal-construction ↑ Prerequisites: place-value
📏 Short solution 💡 3 insights
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Problem
Two three-digit numbers are added. The six digits used in the two numbers are all different. Their sum S is also a three-digit number. Find the smallest possible value of the sum of the digits of S.

Pick an answer.

(A)
1
(B)
4
(C)
5
(D)
15
(E)
20

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The question asks for a minimum, so this is Tool #14 (Extreme Principle): pin down how small the answer can possibly be, then show that small value is actually reachable. Two parts. First, find a hard floor: the two hundreds digits are different and each at least 1, so the sum S can't be too small. That floor lets Tool #3 (Eliminate Possibilities) knock out the digit sums 1, 2, and 3. Second, Tool #6 (Guess and Check) builds one concrete pair of numbers whose sum has digit sum 4, proving 4 is attainable. A bound plus a matching example pins the minimum exactly.

1STEP 1

How small can S be?

Each hundreds digit is at least 1 and the two differ, so they sum to at least 1 + 2 = 3, forcing S ≥ 300.

h₁ ≠ h₂, h₁,h₂ ≥ 1 → h₁ + h₂ ≥ 1 + 2 = 3 → S ≥ 300
2STEP 2

Rule out digit sums 1, 2, 3

Since S ≥ 300, the only digit sum below 4 needs S = 300, but that repeats 0 in a column — banned. So 1, 2, 3 are out.

S ≥ 300, digit sum ≤ 3 → S = 300, but 300 needs repeated 0s → impossible
3STEP 3

Build a sum with digit sum 4

104 + 296 = 400 uses all six digits differently, and 400 has digit sum 4 — reachable, so the minimum is 4. Answer (B).

104 + 296 = 400, 4 + 0 + 0 = 4 = (B)
Answer
4
The example 104 + 296 = 400 checks out by direct addition, its six digits {0,1,2,4,6,9} are all different, and 400 is a three-digit number with digit sum 4. The bound says nothing below 4 is possible, and the example hits exactly 4, so the two halves agree. The losing choices fit the story: (A) 1 is impossible because S ≥ 300; (D) 15 and (E) 20 are far larger than needed.
💡Key takeaway

To make a sum's digits small, use carries to roll columns over to zero — but two different non-zero leading digits keep the sum stuck in the 300s, so 4 is as low as it goes.

  • How small can S be?
  • Rule out digit sums 1, 2, 3
  • Build a sum with digit sum 4