2007 AMC 12B Problem 21

Question

The first

2007

positive integers are each written in base

3.

How many of these base-

3

representations are palindromes? (A palindrome is a number that reads the same forward and backward.)

100

101

102

103

104

Count base-

3

palindromes by their number of digits; note

3^6=729\lt2007\lt2187=3^7

Total all palindromes with up to

7

digits, then remove the

7

-digit ones exceeding

2007=2202100_3

A palindrome is fixed by its first half. Counting base-

3

palindromes by length gives

2

of length

1

or

2,

6

of length

3

or

4,

18

of length

5

or

6,

and

54

of length

7.

That totals

2+2+6+6+18+18

+54=106

palindromes with at most

7

digits. Since

2007=2202100_3,

the

7

-digit palindromes larger than it are

2210122,

2211122,

2212122,

2220222,

2221222,

and

2222222,

which is

6

of them.

Therefore the count is

106-6=100.

Thus, the correct answer is A.

Problem 21 in other years