AMC 8 · 2004 · #8

Grade 6 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

digit-sumdigit-constraintssystematic-enumerationplace-value systematic-enumeration ↑ Prerequisites: place-valuemulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

How many two-digit positive integers have digits that add up to 7?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The unknown is a count, the search space is small, and the digits obey a simple rule (t + u = 7). Tool #2 (Make a Systematic List) lets us walk the tens digit t from 1 up to 7, set u = 7 - t each time, and read off every valid number. Walking t in order guarantees we miss none and repeat none.

1STEP 1

Let t be the tens digit, u the units digit; a two-digit number needs t ≥ 1 with t + u = 7.

t + u = 7, 1 ≤ t ≤ 9, 0 ≤ u ≤ 9

2STEP 2

Since u = 7 - t ≥ 0 forces t ≤ 7, the tens digit t runs from 1 to 7.

u = 7 - t ≥ 0 → t ≤ 7, so t ∈ {1,2,3,4,5,6,7}

3STEP 3

Walk t up in order; each gives one number: 16, 25, 34, 43, 52, 61, 70.

t = 1 → 16, t = 2 → 25, t = 3 → 34, t = 4 → 43, t = 5 → 52, t = 6 → 61, t = 7 → 70

4STEP 4

The list holds 7 numbers, one per value of t, so that is the count.

count = 7 → (B)

Answer

Cross-check by units digit instead. If u = 0, then t = 7 (giving 70); u = 1 → t = 6 (61); ... up to u = 7 → t = 0, but t = 0 is not allowed, so that case drops out. The valid units digits are 0 through 6, which is 7 values. Same answer, reached from the other side — the count is solid.

💡Key takeaway

A clean digit-sum rule plus a systematic list catches every two-digit number once — no Algebra needed, just careful counting from Grade 5 patterns and a Grade 6 bound.

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