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 "1991 AIME Problems/Problem 3"

(→‎Solution)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have
 
Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have
  
Line 16: Line 17:
  
 
In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k_{}^{}=166</math>.
 
In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k_{}^{}=166</math>.
 +
 +
===Solution 2===
 +
 +
We know that once we have found the largest value of <math>k</math>, all values after <math>A_k</math> are less than <math>A_k</math>. Therefore, we are looking for the largest possible value such that:
 +
 +
<math>{\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}</math>
 +
 +
Dividing by <math>{\frac{1}{5}}^k</math> gives:
 +
 +
<math>{1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}</math>.
 +
 +
We can express these binomial coefficients as factorials.
 +
 +
<math>\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}</math>
 +
 +
We note that the <math>1000!</math> can cancel. Also, <math>(1000-k)!=(1000-k)(1000-k-1)!</math>. Similarly, <math>(k+1)!=(k+1)k!</math>.
 +
 +
Canceling these terms yields,
 +
 +
<math>\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}</math>
 +
 +
Cross multiplying gives:
 +
 +
<math>5k+1>1000-k \Rightarrow k>166.5</math>
 +
 +
Therefore, since this identity holds for all values of <math>k>166.5</math>, the largest possible value of <math>k</math> is <math>\boxed{166}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=2|num-a=4}}
 
{{AIME box|year=1991|num-b=2|num-a=4}}

Revision as of 08:33, 15 August 2011

Problem

Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives

${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$. For which $k_{}^{}$ is $A_k^{}$ the largest?

Solution

Solution 1

Let $0<x_{}^{}<1$. Then we may write $A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}$. Taking logarithms in both sides of this last equation and using the well-known fact $\log(a_{}^{}b)=\log a + \log b$ (valid if $a_{}^{},b_{}^{}>0$), we have

$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\log\left[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .$

Now, $\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\frac{(N-j+1)x}{j}>1$ in each of the $\log\big[\;\big]$ terms (recall that $\log y_{}^{} <0$ if $0<y_{}^{}<1$). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=\Big\lfloor\frac{(N+1)x}{1+x}\Big\rfloor$, where $\lfloor z_{}^{}\rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$.

In summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k_{}^{}=166$.

Solution 2

We know that once we have found the largest value of $k$, all values after $A_k$ are less than $A_k$. Therefore, we are looking for the largest possible value such that:

${\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}$

Dividing by ${\frac{1}{5}}^k$ gives:

${1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}$.

We can express these binomial coefficients as factorials.

$\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}$

We note that the $1000!$ can cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$. Similarly, $(k+1)!=(k+1)k!$.

Canceling these terms yields,

$\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}$

Cross multiplying gives:

$5k+1>1000-k \Rightarrow k>166.5$

Therefore, since this identity holds for all values of $k>166.5$, the largest possible value of $k$ is $\boxed{166}$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_3&oldid=41483"