1999 AMC 12 Problem 9

Below is the professionally curated solution for Problem 9 of the 1999 AMC 12, 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.

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Concepts:palindromedistance rate and timebounding to limit cases

Difficulty rating: 1390

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

Solution:

The palindromes after 2979229792 are 29892,29992,30003,29892, 29992, 30003, and 30103.30103. In three hours Ashley can drive at most 375=2253 \cdot 75 = 225 miles.

Reaching 3010330103 would require 3010329792=31130103 - 29792 = 311 miles, which is too far. Reaching 3000330003 requires 3000329792=21130003 - 29792 = 211 miles, giving average speed 2113=7013\dfrac{211}{3} = 70\tfrac13 miles per hour.

Thus, the correct answer is D.

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