AMC 8 · 2002 · #19

Grade 4 countingnumber-theory

Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position

digit-constraintsplace-valuesystematic-enumerationpermutations-basic caseworksystematic-enumeration ↑ Prerequisites: place-valuemulti-digit-arithmetic

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

Count the whole numbers strictly between 99 and 999 that contain the digit 0 exactly once.

Pick an answer.

(A)

(B)

(C)

144

(D)

162

(E)

180

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The hundreds digit cannot be 0, so the single 0 must sit in either the tens slot or the units slot. That observation splits the count into two clean subproblems (Tool #7): "0 in the tens place" and "0 in the units place." Each subproblem is a tiny digit-by-digit choice, perfect for a systematic count (Tool #2). The two subproblems do not overlap (the 0 is in different positions), so we just add the two counts at the end.

1STEP 1

The lead digit h can't be 0, so the lone 0 sits in the tens slot (h 0 u) or the units slot (h t 0) — two disjoint cases.

Total = #{h 0 u} + #{h t 0}

2STEP 2

Case T (h 0 u): h has 9 choices, the tens digit is forced to 0, and u must be nonzero (9 choices), giving 81.

#{h 0 u} = 9 × 1 × 9 = 81

3STEP 3

Case U (h t 0) is symmetric: h has 9 choices, t must be nonzero (9 choices), and the units digit is forced to 0, giving 81.

#{h t 0} = 9 × 9 × 1 = 81

4STEP 4

The two cases never name the same number, so just add them: 81 + 81 = 162.

81 + 81 = 162 → (D)

Answer

162

Sanity-check the size of the answer. There are 9 × 10 × 10 = 900 three-digit numbers total. About

\frac{1}{5}

of them (

\frac{162}{900}

= 0.18) contain exactly one 0, which is reasonable: digits 0 through 9 are equally likely in the tens and units slots, so single-0 numbers should be a meaningful but small slice. Also, 162 is the only choice that equals 2 × 81, which matches the symmetric two-case structure.

💡Key takeaway

The single 0 has only two homes (tens or units), and each home gives 9 × 9 = 81 numbers — so the answer is 81 + 81 = 162.

More like this