Problem:
What is the least positive integer m such that m⋅2!⋅3!⋅4!⋅5!⋯16 ! is a perfect square?
Answer Choices:
A. 30
B. 30,030
C. 70
D. 1430
E. 1001
Solution:
Because n!⋅(n+1)!=n!⋅n!⋅(n+1)=(n!)2⋅(n+1), the given product can be rewritten as
m⋅2⋅(3!)2⋅4⋅(5!)2⋅6⋯(15!)2⋅16
Therefore it suffices to make
m⋅2⋅4⋅6⋅8⋅10⋅12⋅14⋅16=m⋅215⋅32⋅5⋅7
a perfect square. The least possible value of m is 2⋅5⋅7=(C)70.
The problems on this page are the property of the MAA's American Mathematics Competitions