Difference between revisions of "2000 AIME II Problems/Problem 7"
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<cmath>\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.</cmath> | <cmath>\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.</cmath> | ||
− | Recall the [[combinatorial identity| | + | Recall the [[combinatorial identity|Combinatorial Identity]] <math>2^{19} = \sum_{n=0}^{19} {19 \choose n}</math>. Since <math>{19 \choose n} = {19 \choose 19-n}</math>, it follows that <math>\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}</math>. |
Thus, <math>19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. | Thus, <math>19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124</math>. | ||
− | So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>. | + | So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>. |
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+ | == Solution 2 == | ||
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+ | Let <math>f(x) = (1+x)^{19}.</math> Applying the binomial theorem gives us <math>f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.</math> Since <math>\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},</math> <math>N = \frac{2^{18}-20}{19}.</math> After some fairly easy bashing, we get <math>\boxed{137}</math> as the answer. | ||
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+ | ~peelybonehead | ||
+ | |||
+ | ==Solution 3 (Brute Force)== | ||
+ | |||
+ | Convert each denominator to <math>19!</math> and get the numerators to be <math>9,51,204,612,1428,2652,3978,4862</math> (refer to note). Adding these up we have <math>13796</math> therefore <math>\boxed{137}</math> is the desired answer. | ||
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+ | Note: | ||
+ | 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 | ||
== See also == | == See also == |
Latest revision as of 11:09, 5 April 2024
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.