Problem:
What is the greatest integer less than or equal to
396+2963100+2100​?
Answer Choices:
A. 80
B. 81
C. 96
D. 97
E. 625
Solution:
Because the powers-of-3 terms greatly dominate the powers-of-2 terms, the given fraction should be close to
3963100​=34=81
Note that
(3100+2100)−81(396+296)=2100−81⋅296=(16−81)⋅296<0
so the given fraction is less than 81. On the other hand
(3100+2100)−80(396+296)=396(81−80)−296(80−16)=396−2102.
Because 32>23,
396=(32)48>(23)48=2144>2102
it follows that
(3100+2100)−80(396+296)>0
and the given fraction is greater than 80. Therefore the greatest integer less than or equal to the given fraction is (A)80​.
The problems on this page are the property of the MAA's American Mathematics Competitions