Difference between revisions of "2013 AMC 12A Problems/Problem 22"
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Revision as of 16:51, 3 April 2020
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 1
By 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
Solution 2
Let the palindrome be the form in the previous solution which is . It doesn't matter what
is because it only affects the middle digit. There are
ways to pick
and
, and the only answer choice with denominator a factor of
is
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/361
~dolphin7
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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