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

How many whole numbers between 99 and 999 contain exactly one 0?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

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

Givens: The numbers in range are the three-digit integers from $100$ to $998$; Each number has a hundreds digit, a tens digit, and a units digit; Answer choices: (A) $72$, (B) $90$, (C) $144$, (D) $162$, (E) $180$

Unknowns: The count of three-digit numbers whose digits contain exactly one $0$

Understand

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

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List

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.

Execute — Answer: D

#7 Identify Subproblems 3.NBT.A.1 Step 1

Split by where the $0$ sits.
Write the three-digit number as $h\,t\,u$ (hundreds, tens, units).
The hundreds slot $h$ cannot be $0$, so the lone $0$ must be at the tens slot or the units slot.
That gives two non-overlapping cases — Case T ($0$ in the tens place, form $h\,0\,u$) and Case U ($0$ in the units place, form $h\,t\,0$).

$$\text{Total} = \#\{h\,0\,u\} + \#\{h\,t\,0\}$$

💡 Three-digit place value (hundreds, tens, units) is Grade 3; the only restriction is that the lead digit is not $0$.

#2 Make a Systematic List 4.OA.A.3 Step 2

Count Case T: numbers of the form $h\,0\,u$.
The hundreds digit $h$ has $9$ choices ($1$ through $9$).
The tens digit is forced to be $0$, so $1$ choice.
The units digit $u$ must be nonzero (otherwise the number would contain two $0$s), so $9$ choices ($1$ through $9$).
Multiply the per-slot counts.

$$\#\{h\,0\,u\} = 9 \times 1 \times 9 = 81$$

💡 Each independent slot choice multiplies into the total — the same "$9$ shirts $\times$ $9$ pants" reasoning kids meet in Grade 4 word problems.

#2 Make a Systematic List 4.OA.A.3 Step 3

Count Case U: numbers of the form $h\,t\,0$.
The hundreds digit $h$ has $9$ choices ($1$ through $9$).
The tens digit $t$ must be nonzero (or there would be two $0$s), so $9$ choices ($1$ through $9$).
The units digit is forced to be $0$, so $1$ choice.

$$\#\{h\,t\,0\} = 9 \times 9 \times 1 = 81$$

💡 Symmetric to Case T — same counts, just the forced $0$ moved to the units slot.

#7 Identify Subproblems 4.OA.A.3 Step 4

Add the two case counts.
Case T and Case U never describe the same number (the $0$ is in different positions), so the totals just sum.

$$81 + 81 = 162 \;\Rightarrow\; \textbf{(D)}$$

💡 Disjoint cases add. That is the wrap-up move for any "split into subproblems" count.

[1] #7 3.NBT.A.1 Split by where the $0$ sits. Write the three-digit number as $h\,t\,u$ (hundreds

[2] #2 4.OA.A.3 Count Case T: numbers of the form $h\,0\,u$. The hundreds digit $h$ has $9$ choi

[3] #2 4.OA.A.3 Count Case U: numbers of the form $h\,t\,0$. The hundreds digit $h$ has $9$ choi

[4] #7 4.OA.A.3 Add the two case counts. Case T and Case U never describe the same number (the $

Review

Reasonableness: Sanity-check the size of the answer. There are $9 \times 10 \times 10 = 900$ three-digit numbers total. About $\tfrac{1}{5}$ of them ($162/900 = 0.18$) 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 $2 \times 81$, which matches the symmetric two-case structure.

Alternative: Tool #16 (Change Focus / Complement) routed through inclusion-exclusion: count three-digit numbers with $0$ in the tens slot ($9 \cdot 1 \cdot 10 = 90$) plus those with $0$ in the units slot ($9 \cdot 10 \cdot 1 = 90$), then subtract those with $0$ in both ($9 \cdot 1 \cdot 1 = 9$) since they have two $0$s, not one. Result: $90 + 90 - 2 \cdot 9 = 162$. Same answer (D).

CCSS standards used (min grade 4)

3.NBT.A.1 Understand place value of three-digit numbers (hundreds, tens, ones) (Reading each candidate number as a hundreds-tens-units triple $h\,t\,u$ so the single $0$ can be tracked to a specific place.)
4.OA.A.3 Solve multistep word problems with the four operations (Multiplying the per-slot choice counts ($9 \times 1 \times 9 = 81$) within each case, then adding the two cases ($81 + 81 = 162$).)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

More like this