2005 AMC 10A Problem 14

Below is the professionally curated solution for Problem 14 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:digits parity casework

Difficulty rating: 1460

14.

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$41$

$42$

$43$

$44$

$45$

Solution:

The first and last digits must have the same parity so their average is a digit. Both odd gives $5 \cdot 5 = 25$ pairs. Both even, with a nonzero leading digit, gives $4 \cdot 5 = 20$ pairs. Each pair fixes the middle digit, for a total of $25 + 20 = 45$ numbers.

Thus, the correct answer is E.

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