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| − | == Problem 21==
| + | #redirect [[2010 AMC 12B Problems/Problem 11]] |
| − | A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?
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| − | <math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math>
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| − | == Solution ==
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| − | View the palindrome as some number with form (decimal representation):
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| − | <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>\frac{18}{90}=\boxed{\textbf{(E)}\ \frac15}</math>
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| − | ==Another Solution ==
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| − | It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (as it is a four digit palindrome it must be of the form <math>xyyx</math> , where x and y are positive integers from [0,9].
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| − | Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math>
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| − | The <math>98x</math> is now irrelelvant
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| − | Thus we solve:
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| − | <math>11y \equiv 0 \pmod 7</math>
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| − | Which has two solutions: <math>0</math> and <math>7</math>
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| − | There are thus, two options for <math>y</math> out of the 10, so <math>2/10</math> = =\boxed{\textbf{(E)}\ \frac15}$
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| − | == See also ==
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| − | {{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}}
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| − | {{MAA Notice}}
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