Difference between revisions of "2014 AMC 12A Problems/Problem 15"
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\textbf{(E) }45\qquad</math> | \textbf{(E) }45\qquad</math> | ||
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For each digit <math>a=1,2,\ldots,9</math> there are <math>10\cdot10</math> (ways of choosing <math>b</math> and <math>c</math>) palindromes. So the <math>a</math>s contribute <math>(1+2+\cdots+9)(100)(10^4+1)</math> to the sum. | For each digit <math>a=1,2,\ldots,9</math> there are <math>10\cdot10</math> (ways of choosing <math>b</math> and <math>c</math>) palindromes. So the <math>a</math>s contribute <math>(1+2+\cdots+9)(100)(10^4+1)</math> to the sum. | ||
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It just so happens that <cmath> (1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000 </cmath> so the sum of the digits of the sum is <math>18</math>, or <math>\boxed{\textbf{(B)}}</math>. | It just so happens that <cmath> (1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000 </cmath> so the sum of the digits of the sum is <math>18</math>, or <math>\boxed{\textbf{(B)}}</math>. | ||
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==Solution Two== | ==Solution Two== | ||
Revision as of 14:12, 17 March 2015
Problem
A five-digit palindrome is a positive integer with respective digits 
, where 
is non-zero. Let 
be the sum of all five-digit palindromes. What is the sum of the digits of ?
Solution 1
For each digit 
there are 
(ways of choosing 
and 
) palindromes. So the s contribute 
to the sum.
For each digit 
there are 
(since 
) palindromes. So the s contribute 
to the sum.
Similarly, for each 
there are palindromes, so the contributes 
to the sum.
It just so happens that ![\[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\]](http://latex.artofproblemsolving.com/3/8/b/38b77fb7c65e53c50aeee86e0951039d75743e95.png)
so the sum of the digits of the sum is 
, or 
.
Solution Two
Notice that 
In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is 
We have 
palindromes, or 
pairs of palindromes summing to Performing the multiplication gives 
, so the sum is .
Solution Three
As shown above, there are a total of 
five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by to get our sum. The expected value for the ten-thousands and the units digit is 
, and the expected value for the thousands, hundreds, and tens digit is 
. Therefore our expected value is 
. Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either 
or . Thus we only need to calculate 
, and the desired sum is 
.
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
