2007 AMC 12B Problems/Problem 21

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Problem 21

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.)

$\mathrm {(A)}\ 100 \qquad \mathrm {(B)}\ 101 \qquad \mathrm {(C)}\ 102 \qquad \mathrm {(D)}\ 103 \qquad \mathrm {(E)}\ 104$

Solution

$2007_{10} = 2202100_{3}$

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 $52 + 3^3 + 2\cdot 3^2 + 3 = 100 \Rightarrow \mathrm{(A)}$

2007 AMC 12B (Problems • Resources)
Preceded by Problem 20	Followed by Problem 22
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2007 AMC 12B Problems/Problem 21

From AoPSWiki

Problem 21

Solution

See Also