Competition · AMC preparation · step 4 of 4

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)
15
(B)
16
(C)
17
(D)
18
(E)
20

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 120 / d₁, and so on. A quick prime factorization, 120 = 2³ × 3 × 5, makes the divisibility checks instant.

1STEP 1

Factor 120 into primes

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

Pick the biggest first digit

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

Pick the next digit

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

Pick the third digit

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

Fill the last two digits

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

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

8 + 5 + 3 + 1 + 1 = 18 → (D)
Answer
18
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!

  • Factor 120 into primes
  • Pick the biggest first digit
  • Pick the next digit
  • Pick the third digit
  • Fill the last two digits
  • Add the digits

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