2000 AIME II Problems/Problem 7
Problem
Given that
find the greatest integer that is less than .
Solution
Multiplying both sides by yields:
Recall the Combinatorial Identity . Since , it follows that .
Thus, .
So, and .
Solution 2
Let Applying the binomial theorem gives us Since After some fairly easy bashing, we get as the answer.
~peelybonehead
Solution 3 (Brute Force)
Convert each denominator to and get the numerators to be (refer to note). Adding these up we have therefore is the desired answer.
Note: Notice that each numerator is increased each time by a factor of etc. until . If you were taking the test under normal time conditions, it shouldn't be too hard to bash out all of the numbers but it is priority to be careful.
~SirAppel
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.