Difference between revisions of "2013 AMC 12A Problems/Problem 22"
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Revision as of 15:02, 3 July 2013
Problem
A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome 
is chosen uniformly at random. What is the probability that 
is also a palindrome?
Solution
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 
See also
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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 | |
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