Difference between revisions of "2010 AMC 10B Problems/Problem 21"
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Which has two solutions: <math>0</math> and <math>7</math> | Which has two solutions: <math>0</math> and <math>7</math> | ||
| − | There are thus, two options for <math>y</math> out of the 10, so <math>2/10</math> | + | There are thus, two options for <math>y</math> out of the 10, so <math>2/10</math> = \boxed{\textbf{(E)}\ \frac15}$ |
== 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}} | {{MAA Notice}} | ||
Revision as of 22:39, 21 January 2014
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 
Another Solution
It is known that the palindromes can be expressed as: 
(as it is a four digit palindrome it must be of the form 
, where x and y are positive integers from [0,9].
Using the divisibility rules of 7, 
= 
The 
is now irrelelvant
Thus we solve:
Which has two solutions: 
and
There are thus, two options for 
out of the 10, so 
= \boxed{\textbf{(E)}\ \frac15}$
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 | ||
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