1999 AHSME Problems/Problem 9

Problem

Before Ashley started a three-hour drive, her car's odometer reading was 29792, a palindrome. (A palindrome is a number that reads te 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 exceeed the speed limit of 75 miles per hour, which of the following was her greatest possible average speed?

$\mathrm{(A) \ } 33\frac 13 \qquad \mathrm{(B) \ }53\frac 13 \qquad \mathrm{(C) \ }66\frac 23 \qquad \mathrm{(D) \ }70\frac 13 \qquad \mathrm{(E) \ } 74\frac 13$

Solution

Ashley could have traveled at most $3\times 75 = 225$ miles.

Each $5$ -digit palindrome is uniquely determined by its first three digits. The next palindromes are 29892, 29992, 30003, and 30103. We may note that $30103-29792 = 311 > 225$ , so this number and all larger ones are too large.

On the other hand, $30003 - 29792 = 211 \leq 225$ , thus this is the number on the odometer, and the average speed is $\frac{211}3 = \boxed{70 \frac 13}$ .

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1999 AHSME Problems/Problem 9

Problem

Solution

See also