Problem:
Suppose r is a real number for which
⌊r+10019⌋+⌊r+10020⌋+⌊r+10021⌋+⋯+⌊r+10091⌋=546
Find ⌊100r⌋. (For real x,⌊x⌋ is the greatest integer less than or equal to x.)
Solution:
The given sum has 73 terms, each of which equals either ⌊r⌋ or ⌊r⌋+1. This is because 10019,10020,…,10091 are all less than 1. In order for the sum to be 546, it is necessary that ⌊r⌋ be 7, because 73⋅7<546<73⋅8. Now suppose that ⌊r+100k⌋=7 for 19≤k≤m and ⌊r+100k⌋=8 for m+1≤k≤91. Then
7(m−18)+8(91−m)=546,
giving m=56. Thus ⌊r+10056⌋=7 but ⌊r+10057⌋=8. It follows that 7.43≤r<7.44, and hence that ⌊100r⌋=743.
The problems on this page are the property of the MAA's American Mathematics Competitions