AMC12 2007 B

$2007_{10} = 2202100_{3}$ All numbers of six or less digits in base 3 have been written. The form of each palindrome is as follows 1 digit - $a$ 2 digits - $aa$ 3 digits - $aba$ 4 digits - $abba$ 5 digits - $abcba$ 6 digits - $abccba$ Where $a,b,c$ are base 3 digits Since $a \neq 0$, this gives a total of $2 + 2 + 2\cdot 3 + 2\cdot 3 + 2\cdot 3^2 + 2\cdot 3^2 = 52$ palindromes so far. 7 digits - $abcdcba$, but not all of the numbers are less than $2202100_{3}$ Case: $a=1$ All of these numbers are less than $2202100_3$ giving $3^3$ more palindromes Case: $a=2$, $b\neq 2$ All of these numbers are also small enough, giving $2\cdot 3^2$ more palindromes Case: $a=2$, $b=2$ It follows that $c=0$, since any other $c$ would make the value too large. This leaves the number as $220d022_3$. Checking each value of d, all of the three are small enough, so that gives $3$ more palindromes. Summing our cases there are $52 + 3^3 + 2\cdot 3^2 + 3 = 100 \Rightarrow \mathrm{(A)}$