Difference between revisions of "2000 AIME II Problems/Problem 7"
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− | Notice that each numerator is increased each time by a factor of <math>\frac{17}{3}, \frac{16}{4}, \frac{15}{5}, \frac{14}{6},</math> etc. 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. | + | Notice that each numerator is increased each time by a factor of <math>\frac{17}{3}, \frac{16}{4}, \frac{15}{5}, \frac{14}{6},</math> etc. until <math>\frac{11}{9}</math>. 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 | ~SirAppel |
Latest revision as of 12:09, 5 April 2024
Problem
Given that
![$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$](http://latex.artofproblemsolving.com/1/6/d/16d26af464ce4b8d84de3933589ddc93c5d5b57e.png)
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.