1999 AMC 12 Problem 9

Attempt Problem 9 of the 1999 AMC 12 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 1999 AMC 12 solutions, or check the answer key.

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

9.

Before Ashley started a three-hour drive, her car’s odometer reading was 29792,29792, a palindrome. (A palindrome is a number that reads the same way from left to right as it does from right to left.) At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of 7575 miles per hour, which of the following was her greatest possible average speed?
331333\tfrac13
531353\tfrac13
662366\tfrac23
701370\tfrac13
741374\tfrac13
Answer: D
Concepts:palindromedistance rate and timebounding to limit cases
Small hint:
List the palindromes just above 2979229792
Big hint:
The distance in 33 hours cannot exceed 3⋅75=2253 \cdot 75 = 225 miles
Solution:
The palindromes after 2979229792 are 29892,29992,30003,29892, 29992, 30003, and 30103.30103. In three hours Ashley can drive at most 3⋅75=2253 \cdot 75 = 225 miles.
Reaching 3010330103 would require 30103−29792=31130103 - 29792 = 311 miles, which is too far. Reaching 3000330003 requires 30003−29792=21130003 - 29792 = 211 miles, giving average speed 2113=7013\dfrac{211}{3} = 70\tfrac13 miles per hour.
Thus, the correct answer is D.
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.