Difference between revisions of "2010 AMC 10B Problems/Problem 21"
m |
|||
| Line 11: | Line 11: | ||
== See also == | == See also == | ||
{{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:03, 4 July 2013
Problem 21
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 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. 
