2010 AMC 12B Problems/Problem 11
Problem 11
A palindrome between and is chosen at random. What is the probability that it is divisible by ?
Solution
View the palindrome as some number with form (decimal representation): . But because the number is a palindrome, . Recombining this yields . 1001 is divisible by 7, which means that as long as , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 () possibilities for palindromes. However, if , then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |