Difference between revisions of "2010 AMC 10B Problems/Problem 21"
m (→Solution) |
(→Solution) |
||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> ( | + | It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (since it is a four digit palindrome, it must be of the form <math>xyyx</math> , where x and y are integers from [1,9] and [0,9], respectively.) |
Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math> | Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math> | ||
Revision as of 17:03, 22 February 2015
Problem 21
A palindrome between 
and 
is chosen at random. What is the probability that it is divisible by 
?
Solution
It is known that the palindromes can be expressed as: 
(since it is a four digit palindrome, it must be of the form 
, where x and y are integers from [1,9] and [0,9], respectively.)
Using the divisibility rules of 7, 
= 
Since 
, The 
is now irrelelvant.
Thus we solve:
Which has two solutions: 
and
There are thus two options for 
out of the 10, so 
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. 
