Difference between revisions of "2010 AMC 12A Problems/Problem 6"
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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== See also == | == See also == |
Revision as of 20:45, 27 October 2022
- The following problem is from both the 2010 AMC 12A #6 and 2010 AMC 10A #9, so both problems redirect to this page.
Contents
Problem
A , 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
Solution 1
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
.
Solution 2
For to be a four-digit number,
is in between
and
. The palindromes in this range are
,
,
, and
, so the sum of the digits of
can be
,
,
, or
. Only
is an option, and upon checking,
is indeed a palindrome.
Solution 3
Since we know to be
and the only palindrome that works is
, that means
, and so
. So,
+
+
=
.
~songmath20
Video Solution
~Education, the Study of Everything
See also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.