🇺🇸 AMC 8 ⇄ switch contest
2002 AMC 8

Problem 19

Problem 19 · 2002 AMC 8 Hard
Counting & Probability careful-countingplace-value

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

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Answer: D — 162.
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Hint 1 of 2
"Between 99 and 999" just means three-digit numbers — and a 3-digit number can't *start* with 0. So the lone 0 is barred from the hundreds place before you count anything.
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Hint 2 of 2
Place the special digit first: the one 0 must be in the tens or units spot (2 choices), and then make sure the *other* two digits are nonzero so you don't accidentally get a second 0.
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Approach: place the lone 0, then fill the rest with nonzero digits
  1. Deal with the *restricted* digit first — that's the habit. The hundreds digit can't be 0, so the single 0 lives in the tens or units place: 2 spots for it.
  2. The remaining two digits must each be 1–9 (zero would either repeat the 0 or land where it can't): 9 × 9 ways.
  3. Total: 2 × 9 × 9 = 162.
  4. *Why this transfers:* in digit-counting, place the most-constrained slot first (here, where the 0 may go) and keep the "exactly one" rule honest by forbidding the special digit everywhere else — that's what blocks both the leading-zero and the double-counting traps.
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