AMC 10 · 2006 · #19

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesdigit-constraintsdigit-sum casework ↑ Prerequisites: divisibility-rules
📏 Long solution 💡 4 insights
Problem
A four-digit plate uses exactly two digits, each twice, and is divisible by all eight children's ages. Its last two digits give the father's age, and the oldest child is nine. Decide which listed number is not one of the ages.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8
How to solve
Strategy Eliminate Possibilities

The question is 'which one is NOT an age,' so Tool #3 (Eliminate Possibilities) fits: show one candidate is impossible and it must be the answer. Tool #4 (Introduce a Variable) names the two repeated digits a and b, which turns the divisibility-by-9 rule into a clean equation about a+b. Tool #14 (Extreme Principle) says: don't test the candidates blindly — attack the one with the harshest divisibility rule first. A multiple of 5 must end in 0 or 5, the strictest last-digit condition of all the choices, so 5 is the case most likely to break. Chasing that break is exactly what cracks the problem.

1STEP 1

Fix the age set and the digit shape

Nine is always an age, and the doubled digits make the sum even.

ages={1,2,…,9}∖{m}, m∈{1,…,8}, 9 always an age; digit sum=2(a+b)
2STEP 2

Divisibility by 9 forces a + b = 9

That forces the two digits to add to nine.

9 ∣ 2(a+b) → 9 ∣ (a+b), 0 ≤ a+b ≤ 17 → a+b=9
3STEP 3

Test the age 5: it forces the digits to be 0 and 9

Testing one candidate forces the digits to be zero and nine.

5 ∣ N and 2 ∣ N → N ends in 0 → {a,b}={0,9}, digits 0,0,9,9
4STEP 4

No plate of 0,0,9,9 works, so 5 is impossible

Neither arrangement works, so that candidate is impossible.

{9900, 9090}: 9900 → age 00 (no), 4 ∤ 9090 (no) → 5 is not an age → (B)
5STEP 5

Confirm 5 is the missing age with a real plate

A real plate confirms it, so the answer is 5, choice (B).

lcm(1,2,3,4,6,7,8,9)=504, 504 × 11=5544 (digits 5,5,4,4; age 44)
Answer
5
The logic hangs on two unbreakable rules. Divisibility by 9 (guaranteed, since a 9-year-old exists) forces the two digits to add to 9. Divisibility by 5 would force the digits to be 0 and 9, and no arrangement of 0,0,9,9 can be a real plate — it either makes the father 0 years old or fails divisibility by 4. Meanwhile the concrete plate 5544 shows the leave-out-5 world is perfectly consistent: 5544=504 × 11 is divisible by 1,2,3,4,6,7,8,9 but not by 5. Everything points to 5.
💡Key takeaway

A 9-year-old forces the two repeated digits to add to 9, and a 5 would force the digits to be 0 and 9 — which can never make a real plate — so 5 is the age nobody has, and 5544 proves it.

  • Fix the age set and the digit shape
  • Divisibility by 9 forces a + b = 9
  • Test the age 5: it forces the digits to be 0 and 9
  • No plate of 0,0,9,9 works, so 5 is impossible
  • Confirm 5 is the missing age with a real plate

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