AMC 10 · 2006 · #25

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
Mr. Jones has eight children, all different ages, and the oldest is 9. A 4-digit license plate uses exactly two different digits, each appearing twice (for example a number shaped like aabb or abab). This number is divisible by every one of the eight children's ages, and its last two digits equal Mr. Jones's own age. Among the choices 4,5,6,7,8, decide which one is not the age of any of his children.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Eight distinct ages at most 9 must be 1 through 9 with one value dropped, and 9 stays. The repeated digits a and b give digit sum 2(a+b).

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

The plate is a multiple of 9, so 2(a+b) is too; 2 shares no factor with 9 and a+b is at most 17, so a+b=9.

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

If 5 were an age the plate would be a multiple of 10, so 0 repeats and its partner is 9: the digits are 0,0,9,9.

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

Only 9900 and 9090 fit, but 9900 makes the father 00 and 9090 is not a multiple of 4, so the missing age is 5.

{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

Dropping 5 leaves 1,2,3,4,6,7,8,9, whose lcm is 504, and 504 × 11 = 5544 fits: two digits twice each, father aged 44.

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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