Problem:
Find the least positive integer k for which the equation ⌊n2002⌋=k has no integer solutions for n. (The notation ⌊x⌋ means the greatest integer less than or equal to x.)
Solution:
The equation ⌊n2002⌋=k is equivalent to
k≤n2002<k+1, or k+12002<n≤k2002
In order that there be no solutions, there can be no integer in this interval, that is, k2002 and k+12002 must have the same integer part. The length of the interval must be less than 1, so
k2002−k+12002<1
which yields k(k+1)>2002, and thus k≥45. For k=45,46,47,48,49,50, the integer part of 2002/k is 44,43,42,41,40,40, respectively. Thus 2002/49 and 2002/50 have the same integer part, so the least positive integer value of k is 49.
The problems on this page are the property of the MAA's American Mathematics Competitions