Difference between revisions of "2010 AMC 10A Problems/Problem 9"
| Line 11: | Line 11: | ||
== See also == | == See also == | ||
{{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}} | {{AMC10 box|year=2010|num-b=8|num-a=10|ab=A}} | ||
| − | + | {{AMC12 box|year=2010|num-b=5|num-a=7|ab=A}} | |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:13, 10 May 2020
Problem
A palindrome, such as 
, is a number that remains the same when its digits are reversed. The numbers 
and 
are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?
Solution
is at most 
, so is at most 
. The minimum value of is 
. However, the only palindrome between and 
is 
, which means that must be .
It follows that is 
, so the sum of the digits is 
.
See also
| 2010 AMC 10A (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 | ||
| All AMC 10 Problems and Solutions | ||
| 2010 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
