Difference between revisions of "2007 AMC 12B Problems/Problem 21"
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
The first <math>2007</math> positive integers are each written in base <math>3</math>. How many of these base-<math>3</math> representations are palindromes? (A palindrome is a number that reads the same forward and backward.) | The first <math>2007</math> positive integers are each written in base <math>3</math>. How many of these base-<math>3</math> representations are palindromes? (A palindrome is a number that reads the same forward and backward.) | ||
Latest revision as of 16:35, 15 February 2021
Problem
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.