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.
Problem
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 
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 b (0-9), 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
