2010 AMC 12B Problems/Problem 11
- The following problem is from both the 2010 AMC 12B #11 and 2010 AMC 10B #21, so both problems redirect to this page.
Contents
Problem
A palindrome between and is chosen at random. What is the probability that it is divisible by ?
Solution 1
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
Solution (Divisibility Rules)
We can notice the palindrome is of the form . Then, by the divisibility rule of , must divide This nicely simplifies to the fact that so is clearly or . This gives us total choices for the palindrome divisible by , divided by total choices for , giving us an answer of .
~icecreamrolls8
Addendum (Alternate)
and . Knowing that does not factor (pun intended) into the problem, note 110's prime factorization and . There are only 10 possible digits for , 0 through 9, but only holds if . This is 2 of the 10 digits, so
~BJHHar
Video Solution
~IceMatrix
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 |
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 10 Problems and Solutions |
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