Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME II Problems/Problem 7"

(\)
 
(→‎Solution 3 (Brute Force))
 
(15 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Given that <center><math>\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!}</math></center> find the [[floor function|greatest integer]] that is less than <math>\frac N{100}</math>.
  
 
== Solution ==
 
== Solution ==
 +
Multiplying both sides by <math>19!</math> yields:
 +
 +
<cmath>\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.</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|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>.
 +
 +
So, <math>N=\frac{262124}{19}=13796</math> and <math>\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}</math>.
 +
 +
 +
== Solution 2 ==
 +
 +
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.
 +
 +
~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.
 +
 +
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 ==
−
* [[2000 AIME II Problems]]
+
{{AIME box|year=2000|n=II|num-b=6|num-a=8}}
 +
 
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 11:09, 5 April 2024

Problem

Given that

$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$

find the greatest integer that is less than $\frac N{100}$.

Solution

Multiplying both sides by $19!$ yields:

\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]

\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.\]

Recall the Combinatorial Identity $2^{19} = \sum_{n=0}^{19} {19 \choose n}$. Since ${19 \choose n} = {19 \choose 19-n}$, it follows that $\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}$.

Thus, $19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124$.

So, $N=\frac{262124}{19}=13796$ and $\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}$.


Solution 2

Let $f(x) = (1+x)^{19}.$ Applying the binomial theorem gives us $f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.$ Since $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{18}-20}{19}.$ After some fairly easy bashing, we get $\boxed{137}$ as the answer.

~peelybonehead

Solution 3 (Brute Force)

Convert each denominator to $19!$ and get the numerators to be $9,51,204,612,1428,2652,3978,4862$ (refer to note). Adding these up we have $13796$ therefore $\boxed{137}$ is the desired answer.

Note: Notice that each numerator is increased each time by a factor of $\frac{17}{3}, \frac{16}{4}, \frac{15}{5}, \frac{14}{6},$ etc. until $\frac{11}{9}$. 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_7&oldid=217857"