Difference between revisions of "2007 AMC 12B Problems/Problem 21"
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Revision as of 09:52, 4 July 2013
Problem 21
The first 
positive integers are each written in base 
. How many of these base- representations are palindromes? (A palindrome is a number that reads the same forward and backward.)
Solution
All numbers of six or less digits in base 3 have been written.
The form of each palindrome is as follows
1 digit - 
2 digits - 
3 digits - 
4 digits - 
5 digits - 
6 digits - 
Where 
are base 3 digits
Since 
, this gives a total of 
palindromes so far.
7 digits - 
, but not all of the numbers are less than 
Case: 
All of these numbers are less than 
giving 
more palindromes
Case: 
, 
All of these numbers are also small enough, giving 
more palindromes
Case: , 
It follows that 
, since any other 
would make the value too large. This leaves the number as 
. Checking each value of d, all of the three are small enough, so that gives more palindromes.
Summing our cases there are
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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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