Problem:
The number 2013 is expressed in the form
2013=b1!b2!⋯bn!a1!a2!⋯am!
where a1≥a2≥⋯≥am and b1≥b2≥⋯≥bn are positive integers and a1+b1 is as small as possible. What is ∣a1−b1∣?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
The prime factorization of 2013 is 3⋅11⋅61. There must be a factor of 61 in the numerator, so a1≥61. Since a1! will have a factor of 59 and 2013 does not, there must be a factor of 59 in the denominator, and b1≥59. Thus a1+b1≥120, and this minimum value can be achieved only if a1=61 and b1=59. Furthermore, this minimum value is attainable because
2013=(59!)(10!)(5!)(61!)(11!)(3!)
Thus ∣a1−b1∣=a1−b1=61−59=(B)2.
The problems on this page are the property of the MAA's American Mathematics Competitions