Difference between revisions of "2009 AMC 10B Problems/Problem 11"
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== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2009|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:32, 28 June 2021
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 
. There are 
permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. 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. 
