Difference between revisions of "2009 AMC 10B Problems/Problem 11"
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This means that our answer is <math>2+2+2=\boxed{\textbf{(A)}6}</math> | This means that our answer is <math>2+2+2=\boxed{\textbf{(A)}6}</math> | ||
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== See Also == | == See Also == | ||
Revision as of 13:17, 28 September 2020
Contents
Problem
How many 
-digit palindromes (numbers that read the same backward as forward) can be formed using the digits 
, , 
, , 
, , ?
Solution
A seven-digit palindrome is a number of the form 
. Clearly, 
must be , as we have an odd number of fives. We are then left with 
. Each of the 
permutations of the set 
will give us one palindrome.
Solution 2
Say we have a 2 first. Then, we have a 2 pinned as the last digit, so we have to fill in the remaining digits with only 3's and 5's. We have 2 options for the second digit then, and the rest is fixed. This means that we have ways for this case.
Say we have a 3 first. By symmetry, this is the same as the 2 cases, so we have ways.
Say we have a 5 first. We then have a 5 in the middle. We can either have a 2 second or a 3 second. So we have ways.
This means that our answer is 
~yofro
See Also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
