Difference between revisions of "1999 AHSME Problems/Problem 9"
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Each <math>5</math>-digit palindrome is uniquely determined by its first three digits. The next palindromes are 29892, 29992, 30003, and 30103. We may note that <math>30103-29792 = 311 > 225</math>, so this number and all larger ones are too large. | Each <math>5</math>-digit palindrome is uniquely determined by its first three digits. The next palindromes are 29892, 29992, 30003, and 30103. We may note that <math>30103-29792 = 311 > 225</math>, so this number and all larger ones are too large. | ||
| − | On the other hand, <math>30003 - 29792 = 211 \leq 225</math>, thus this is the number on the odometer, and the average speed is <math>\frac{211}3 = \boxed{(D) 70 \frac 13}</math>. | + | On the other hand, <math>30003 - 29792 = 211 \leq 225</math>, thus this is the number on the odometer, and the average speed is <math>\frac{211}3 = \boxed{\textbf{(D) 70 \frac 13}}</math>. |
== See also == | == See also == | ||
{{AHSME box|year=1999|num-b=8|num-a=10}} | {{AHSME box|year=1999|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:34, 21 May 2021
Problem
Before Ashley started a three-hour drive, her car's odometer reading was 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 75 miles per hour, which of the following was her greatest possible average speed?
Solution
Ashley could have traveled at most 
miles.
Each 
-digit palindrome is uniquely determined by its first three digits. The next palindromes are 29892, 29992, 30003, and 30103. We may note that 
, so this number and all larger ones are too large.
On the other hand, 
, thus this is the number on the odometer, and the average speed is $\frac{211}3 = \boxed{\textbf{(D) 70 \frac 13}}$ (Error compiling LaTeX. Unknown error_msg).
See also
| 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. 
