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| − | ==Problem==
| + | #REDIRECT [[2013 AMC 12B Problems/Problem 15]] |
| − | The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> where <math> a_1\ge a_2\ge\cdots\ge a_m </math> and <math> b_1\ge b_2\ge\cdots\ge b_n </math> are positive integers and <math> a_1+b_1 </math> is as small as possible. What is <math>|a_1-b_1|</math>?
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| − | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}}\ 4\qquad\textbf{(E)}\ 5 </math>
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| − | ==Solution==
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| − | The prime factorization of <math> 2013 </math> is <math> 61*11*3 </math>. To have a factor of <math>61</math> in the numerator, <math>a_1</math> must equal <math>61</math>. Now we notice that there can be no prime <math>p</math> which is not a factor of 2013 such that <math> b_1<p<61</math> because this prime will not be represented in the denominator, but will be represented in the numerator. The highest <math> p </math> less than <math>61</math> is <math>59</math>, so there must be a factor of <math>59</math> in the denominator. It follows that <math>b_1 = 59</math>, so the answer is <math>|61-59|</math>, which is <math>\boxed{\textbf{(B) }2}</math>. One possible way to express <math> 2013 </math> is <cmath> \frac{61!*19!*11!}{59!*20!*10!}, </cmath>
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