Problem 15 · 1985 AJHSME
Hard
Counting & Probability
complementary-countingdigits
How many whole numbers between 100 and 400 contain the digit 2?
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Answer: C — 138.
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Hint 1 of 2
Counting numbers that contain a 2 directly is messy — a 2 might be in the hundreds, tens, or units place, and numbers like 220 or 282 would get counted twice. When 'at least one' is hard, count the OPPOSITE.
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Hint 2 of 2
Count how many numbers have NO 2 at all, then subtract from the total. 'Total − (none) = (at least one)' sidesteps all the double-counting. The numbers run 100 to 399, so there are 300 of them.
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Approach: complement count
- There are 300 whole numbers from 100 to 399. Instead of counting those that contain a 2, count those that contain NO 2 (much cleaner — no overlaps).
- No-2 count: the hundreds digit can be 1 or 3 (2 ways, since 2 is banned and a 3-digit number can't start with 0), and each of the tens and units can be any digit except 2 (9 ways each): 2 × 9 × 9 = 162.
- Contains a 2 = total − none = 300 − 162 = 138.
- Why this transfers: whenever you want 'at least one' of something, counting the cases with NONE and subtracting is almost always easier — it turns an overlapping mess into one clean multiplication.
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