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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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| − | 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 <math>[1, 9]</math> and <math>[0, 9]</math>, respectively.)
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| − | We simplify this to:
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| − | <math>1001x+110y</math>.
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| − | Because the question asks for it to be divisible by 7,
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| − | We express it as <math>1001x+110y \equiv 0 \pmod 7</math>.
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| − | Because <math>1001 \equiv 0 \pmod 7</math>,
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| − | We can substitute <math>1001</math> for <math>0</math>
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| − | We are left with <math>110y \equiv 0 \pmod 7</math>
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| − | Since <math>110 \equiv 5 \pmod 7</math> we can simplify the <math>110</math> in the expression to
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| − | <math>5y \equiv 0 \pmod 7</math>.
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| − | In order for this to be true, <math>y \equiv 0 \pmod 7</math> must also be true.
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| − | Thus we solve:
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| − | <math>y \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 the answer is <math>2/10 = \boxed{\textbf{(E)}\ \frac15}</math>
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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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