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| − | == Problem ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 1]] |
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| − | The arithmetic mean of the nine numbers in the set <math>\{9,99,999,9999,...,999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit
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| − | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 8 </math>
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| − | == Solution ==
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| − | We wish to find <math>\frac{9+99+\cdots +999999999}{9}</math>, or <math>\frac{9(1+11+111+\cdots +111111111)}{9}=123456789</math>. This does not have the digit 0, so <math>\mathrm{ (A) \ }</math>
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}}
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| − | [[Category:Introductory Algebra Problems]] | |
