Difference between revisions of "2010 AMC 12B Problems/Problem 11"
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− | <math>7\mid 1001a^3+110b^2</math> and <math>1001 \equiv 0 \pmod 7</math>. Knowing that <math>a</math> does not factor (pun intended) into the problem, note 110's prime factorization and <math>7\mid b</math>. There are only 10 possible digits for b (0-9), but <math>7\mid b</math> only holds if <math>b=0, 7</math>. This is 2 of the 10 digits, so <math>\frac{2}{10}=\boxed{\textbf{E)}\frac{1}{5}}</math> | + | <math>7\mid 1001a^3+110b^2</math> and <math>1001 \equiv 0 \pmod 7</math>. Knowing that <math>a</math> does not factor (pun intended) into the problem, note 110's prime factorization and <math>7\mid b</math>. There are only 10 possible digits for <math>b (0-9)</math>, but <math>7\mid b</math> only holds if <math>b=0, 7</math>. This is 2 of the 10 digits, so <math>\frac{2}{10}=\boxed{\textbf{E)}\frac{1}{5}}</math> |
~BJHHar | ~BJHHar |
Revision as of 15:42, 11 July 2021
- 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 , 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.