2013 AMC 12A Problems/Problem 22
Revision as of 05:54, 7 February 2013 by Epicwisdom (talk | contribs) (Created page with "Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: <math>A (A+B) (B+C) (B+C) (A+B) A</math> Note that if...")
Working backwards, we can multiply 5-digit palindromes by
, giving a 6-digit palindrome:
Note that if or
, then the symmetry will be broken by carried 1s
Simply count the combinations of for which
and
implies
possible
(0 through 8), for each of which there are
possible C, respectively. There are
valid palindromes when
implies
possible
(0 through 7), for each of which there are
possible C, respectively. There are
valid palindromes when
Following this pattern, the total is
6-digit palindromes are of the form , and the first digit cannot be a zero, so there are
combinations of
So, the probability is