2013 AMC 12A Problems/Problem 22
Contents
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
Note
You can more easily count the number of triples by noticing that there are possible values for and possible values for once is chosen. Summing over all , the number is By the hockey-stick identity, it is .
~rayfish
Solution 2 (using the answer choices)
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 .
Solution 3 (extremely simple)
A six-digit palindrome, that when divided by 11, has to, if another palindrome, equal a five-digit palindrome. It cannot be a four-digit palindrome, otherwise, it would be too small. We now have a five-digit palindrome , where the digits could be the same. If we multiply by we are adding like this:
ABCBA0
+ ABCBA=
A, A+B, B+C, B+C, A+B, A
Commas separating the digits, We don't have to worry about A because it can be any digit, however, we have to do casework on B, as it affects all the digits.
B=0 A=1-9 C=0-9, B=1 A=1-8 C=0-8, B=2 A=1-7 C=0-7, B=3 A=1-6 C=0-6, B=4 A=1-5 C=0-5, B=5 A=1-4 C=0-4, B=6 A=1-3 C=0-3, B=7 A=1-2 C=0-2, B=8 A=1 C=0-1,
B=9, does not work because A's only option is equal to 0 which makes it an invalid number (four-digit)
These solutions work as we ensure that all the digits do not carry into the following digits (e.g., The tens cannot carry into the hundreds or it would not be a palindrome).
For probability, as Richard Rusczyk says the total possibilities over the total successful outcomes Possibilities--> which are A=1-9, B=0-9, and C=0-9 --> 9 x 10 x 10 = 900
Successful Outcomes--> which are from the casework above --> =
330/900 =
~Solution by Daily Dose of Math (Thesmartgreekmathdude)
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 |
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