AMC 10 · 2003 · #13

Grade 2 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
digit-sumsystematic-enumeration work-backwardscasework ↑ Prerequisites: digit-sum
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
The symbol ♣(x) means add up the digits of x; for example ♣(8)=8 and ♣(123)=1+2+3=6. So ♣(♣(x)) means applying that digit sum twice — first to x, then to the result. Here x runs over the two-digit numbers, 10 through 99. For how many such x is ♣(♣(x)) equal to 3?

Pick an answer.

(A)
3
(B)
4
(C)
6
(D)
9
(E)
10

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Work Backwards

The condition is stacked: the outer ♣ acts on the inner one. So work backwards (Tool #11) — name the inner digit sum s = ♣(x) and first ask which values of s make ♣(s) = 3. To keep that search finite, pin down the range of s with a boundary check (Tool #14): a two-digit number's digits can sum to at most 9+9=18, so s lives in 1 to 18. That turns the outer condition into just two allowed inner sums, s=3 and s=12. Then the problem splits into two independent counting subproblems (Tool #7), each solved by a short systematic list (Tool #2): count two-digit numbers whose digits add to 3, and those whose digits add to 12. Add the two counts.

1STEP 1

Name the inner sum and bound it

Let s = ♣(x). A two-digit number's digits add to at least 1+0=1 and at most 9+9=18, so 1 ≤ s ≤ 18.

s = ♣(x), 1 ≤ s ≤ 18
2STEP 2

Find which inner sums work

Now ♣(s)=3. Within 1–18 a one-digit s must be 3, and a two-digit s must have digits summing to 3, so s = 3 or s = 12.

♣(s)=3 with 1 ≤ s ≤ 18 → s = 3 or s = 12
3STEP 3

Count two-digit numbers with digit sum 3

Write x as tens digit a plus ones digit b with a+b=3. Then a=1,2,3 give 12, 21, 30, and a=4 would need b=-1 — 3 numbers.

a+b=3: 12, 21, 30 → 3 numbers
4STEP 4

Count two-digit numbers with digit sum 12

For a+b=12, b=12-a must stay in 0–9, so a runs from 3 to 9: 39, 48, 57, 66, 75, 84, 93 — 7 numbers.

a+b=12: 39, 48, 57, 66, 75, 84, 93 → 7 numbers
5STEP 5

Add the two counts

A number has one digit sum, so the two cases never overlap: 3 + 7 = 10 two-digit numbers, choice (E).

3 + 7 = 10 → (E)
Answer
10
The count 10 is comfortably inside what's possible: there are 90 two-digit numbers, and requiring a specific double digit-sum should keep only a small handful, so 10 is reasonable. Spot-check the two edge members: 12 → ♣(12)=3 → ♣(3)=3 ✓, and 93 → ♣(93)=12 → ♣(12)=3 ✓. Both lists are also self-consistent: digit sums 3 and 12 are exactly the values in 1–18 whose digits add to 3, so nothing was missed and nothing extra slipped in.
💡Key takeaway

Work from the outside in: figure out what the first digit sum must be (3 or 12), then list the two-digit numbers that hit each and add the counts.

  • Name the inner sum and bound it
  • Find which inner sums work
  • Count two-digit numbers with digit sum 3
  • Count two-digit numbers with digit sum 12
  • Add the two counts

More like this

Same archetype — closest grade level first.