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=2​.
The problems on this page are the property of the MAA's American Mathematics Competitions