2004 AMC 10B Problem 2 — Solution | LIVE by Po-Shen Loh

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2004 AMC 10B Problem 2

Attempt Problem 2 of the 2004 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

How many two-digit positive integers have at least one $7$ as a digit?

$10$

$18$

$19$

$20$

$30$

Answer: B

Concepts:digits inclusion-exclusion

Difficulty rating: 920

Solution:

The numbers $70$ through $79$ give $10$ with a $7$ in the tens place.

The numbers $17, 27, \ldots, 97$ give $9$ with a $7$ in the units place.

Since $77$ is counted twice, the total is $10 + 9 - 1 = 18.$

Thus, the correct answer is B.

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Problem 2 in Other Years

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