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

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2004 AMC 10B Problem 2

Below is the professionally curated solution for Problem 2 of the 2004 AMC 10B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:digits inclusion-exclusion

Difficulty rating: 920

2.

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

$10$

$18$

$19$

$20$

$30$

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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