AMC 10 · 2005 · #8

Grade 6 number-theory
prime-factorizationdigit-constraintsplace-value bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: prime-factorizationplace-valuefactors
📏 Medium solution 💡 2 insights
Problem
Three digits build both a three-digit number and their own digit sum. The product of those two is exactly 2005. Determine the leading digit.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Eliminate Possibilities

Searching all 1000 digit triples is hopeless by hand, but the equation is a product of two whole numbers, and 2005 has very few divisors. So list every divisor of 2005, use the largest possible digit values to cap each factor, and cross out every pairing that breaks a cap. If only one pairing survives, the digits are forced — which is stronger than merely exhibiting a triple that happens to work.

1STEP 1

Name the two factors

Both factors are whole, so each is a divisor of the product.

N = 100A + 10M + C, S = A + M + C, N · S = 2005
2STEP 2

Factor 2005 completely

Factoring completely leaves only four divisors.

2005 = 5 · 401, divisors: 1, 5, 401, 2005
3STEP 3

Cap each factor using the digit limit

Maximum digits cap the two factors at 999 and 27.

N ≤ 999 and S ≤ 27
4STEP 4

Cross out every impossible pairing

Only one pairing survives both caps.

N ∈ {1, 5, 401}, S ∈ {1, 5} ⟹ (N, S) = (401, 5)
5STEP 5

Read the digits back out of 401

Place value reads the digits back, so the leading digit is 4, choice (D).

100A + 10M + C = 401 ⟹ (A, M, C) = (4, 0, 1), 401 · 5 = 2005
Answer
4
Substitute the digits back into the original equation: (100 · 4 + 10 · 0 + 1)(4 + 0 + 1) = 401 · 5 = 2005, exactly as required. The answer is also unique rather than merely consistent: the elimination in the fourth step examined every divisor pairing allowed by the digit caps and only (401, 5) survived, and place value then pinned (A, M, C) = (4, 0, 1) with no freedom left. So no other digit triple can satisfy the equation, and the other four answer choices are impossible.
💡Key takeaway

When two whole numbers multiply to a fixed value, list that value's divisors first — then the size limits on each factor usually leave only one choice.

  • Name the two factors
  • Factor 2005 completely
  • Cap each factor using the digit limit
  • Cross out every impossible pairing
  • Read the digits back out of 401