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