Difference between revisions of "2013 AMC 12A Problems/Problem 22"
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...") |
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+ | == 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 <math>n</math> is chosen uniformly at random. What is the probability that <math>\frac{n}{11}</math> is also a palindrome? | ||
+ | |||
+ | <math> \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: | Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: | ||
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So, the probability is <math>\frac{330}{900} = \frac{11}{30}</math> | So, the probability is <math>\frac{330}{900} = \frac{11}{30}</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=21|num-a=23}} |
Revision as of 18:54, 22 February 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 |