2005 AMC 10A Problem 14

Attempt Problem 14 of the 2005 AMC 10A 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 2005 AMC 10A solutions, or check the answer key.

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

14.

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
4141
4242
4343
4444
4545
Answer: E
Concepts:digitsparitycasework
Small hint:
The middle digit is an integer only when the first and last digits share the same parity
Big hint:
Count first-and-last digit pairs that are both odd or both even, remembering the first digit cannot be 00
Solution:
The first and last digits must have the same parity so their average is a digit. Both odd gives 5⋅5=255 \cdot 5 = 25 pairs. Both even, with a nonzero leading digit, gives 4⋅5=204 \cdot 5 = 20 pairs. Each pair fixes the middle digit, for a total of 25+20=4525 + 20 = 45 numbers.
Thus, the correct answer is E.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.