Difference between revisions of "2014 AMC 12A Problems/Problem 15"
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==Solution Two== | ==Solution Two== | ||
− | As there are only <math>9\cdot10\cdot10 = 900</math> five digit palindromes, it is sufficient to add up all of them. <math></math>. | + | As there are only <math>9\cdot10\cdot10 = 900</math> five digit palindromes, it is sufficient to add up all of them. <math>a</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:49, 9 February 2014
Contents
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 One
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 so the sum of the digits of the sum is , or .
(Solution by AwesomeToad)
Solution Two
As there are only five digit palindromes, it is sufficient to add up all of them. .
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 |
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