AMC 8 · 2018 · #14

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

prime-factorizationfactorsdigit-constraintsplace-valuedigit-sum systematic-enumerationcasework ↑ Prerequisites: prime-factorizationplace-value

📏 Medium solution 💡 3 insights

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Problem

Among all five-digit numbers whose five digits multiply to 120, find the greatest one (call it N), and report the sum of its digits.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

To maximize a multi-digit number, the leftmost (highest place value) digit matters most, then the next, and so on. So we use Tool #6 (Guess and Check) one digit position at a time: try the biggest candidate (9, then 8, then 7, …) and keep the first one that still lets the remaining digits multiply to the leftover product. Tool #7 (Identify Subproblems) splits the whole task into five smaller "pick one digit" problems: after we choose d₁, we just need four digits whose product is $\frac{120}{d₁}$ , and so on. A quick prime factorization, 120 = 2³ × 3 × 5, makes the divisibility checks instant.

1STEP 1

Prime-factorize 120 = 2³ × 3 × 5, so the only single-digit factors are {1, 2, 3, 4, 5, 6, 8} — note 7 and 9 don't divide 120.

120 = 2³ × 3 × 5

2STEP 2

Take the largest workable ten-thousands digit: 9 fails, but 120 ÷ 8 = 15 works, so d₁ = 8 and the remaining four digits must multiply to 15.

d₁ = 8, 120 ÷ 8 = 15

3STEP 3

For the thousands digit, none of 9–6 divides 15, but 5 does (15 ÷ 5 = 3), so d₂ = 5 and the remaining three digits must multiply to 3.

d₂ = 5, 15 ÷ 5 = 3

4STEP 4

The largest single-digit factor of the remaining product 3 is 3 itself, so d₃ = 3 and the last two digits must multiply to 3 ÷ 3 = 1.

d₃ = 3, 3 ÷ 3 = 1

5STEP 5

The last two digits multiply to 1, so d₄ = d₅ = 1; arranging {8, 5, 3, 1, 1} biggest-first gives N = 85311.

d₄ = 1, d₅ = 1 → N = 85311

6STEP 6

Add the digits of N: 8 + 5 + 3 + 1 + 1 = 18, which is choice (D).

8 + 5 + 3 + 1 + 1 = 18 → (D)

Answer

Verify the product first: 8 × 5 × 3 × 1 × 1 = 40 × 3 = 120. Good. Could any larger five-digit number N' ＞ 85311 still have digit product 120? Any such N' would need a ten-thousands digit of 9, but 9 does not divide 120, so the four remaining digits could not multiply to a whole number — impossible. So 85311 really is the greatest, and the digit sum 8+5+3+1+1 = 18 matches choice (D).

💡Key takeaway

This AMC 8 problem only needs Grade 4 factor pairs and place value you already know!

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