Difference between revisions of "2010 AMC 12A Problems/Problem 6"
Line 14: | Line 14: | ||
===Solution 2=== | ===Solution 2=== | ||
For <math>x+32</math> to be a four-digit number, <math>x</math> is in between <math>968</math> and <math>999</math>. The palindromes in this range are <math>969</math>, <math>979</math>, <math>989</math>, and <math>999</math>, so the sum of digits of <math>x</math> can be <math>24</math>, <math>25</math>, <math>26</math>, or <math>27</math>. Only <math>\boxed{\textbf{(E)}\ 24}</math> is an option, and upon checking, <math>x+32=1001</math> is indeed a palindrome. | For <math>x+32</math> to be a four-digit number, <math>x</math> is in between <math>968</math> and <math>999</math>. The palindromes in this range are <math>969</math>, <math>979</math>, <math>989</math>, and <math>999</math>, so the sum of digits of <math>x</math> can be <math>24</math>, <math>25</math>, <math>26</math>, or <math>27</math>. Only <math>\boxed{\textbf{(E)}\ 24}</math> is an option, and upon checking, <math>x+32=1001</math> is indeed a palindrome. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=P7rGLXp_6es | ||
+ | ~IceMatrix | ||
== See also == | == See also == |
Revision as of 14:00, 26 May 2020
- The following problem is from both the 2010 AMC 12A #6 and 2010 AMC 10A #9, so both problems redirect to this page.
Problem
A , such as 83438, 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 digits of
can be
,
,
, or
. Only
is an option, and upon checking,
is indeed a palindrome.
Video Solution
https://www.youtube.com/watch?v=P7rGLXp_6es ~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.