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AMC 8 · 2002 · #19

Easy mode Grade 4
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Problem

Picture all the whole numbers between 99 and 999. These are the three-digit numbers — 100, 101, 102, all the way up to 998.

Look at the three digits of each number. We want the ones that have exactly one 0 among their digits — not zero zeros, not two zeros, exactly one.

For example, 105 counts (it has one 0) and 270 counts (it has one 0). But 100 does not count (it has two 0s), and 234 does not count (it has no 0s).

How many three-digit numbers have exactly one 0?

Pick an answer.

(A)
72
(B)
90
(C)
144
(D)
162
(E)
180
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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.

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) $180$

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 $81 + 81 = 162$.

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

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