Difference between revisions of "2010 AMC 12B Problems/Problem 11"
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== Solution == | == Solution == | ||
+ | == Solution == | ||
+ | View the palindrome as some number with form (decimal representation): | ||
+ | <math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, 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 <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} |
Revision as of 20:54, 4 September 2010
Contents
Problem 11
A palindrome between and is chosen at random. What is the probability that it is divisible by ?
Solution
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 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |