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Difference between revisions of "1991 AIME Problems/Problem 3"

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<math>
 
<math>
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\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)}{j}x\right]
+
\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)}{j}x\right]\, .
 
</math>
 
</math>
 +
 +
Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)}{j}x>1</math> in each of the terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the <math>k_{}^{}</math> that we are looking for must satisfy <math>\frac{(N-k+1)}{k}x>1</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 20:13, 20 April 2007

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

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

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

Now, $\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\frac{(N-j+1)}{j}x>1$ in each of the terms (recall that $\log y_{}^{} <0$ if $0<y_{}^{}<1$). Therefore, the $k_{}^{}$ that we are looking for must satisfy $\frac{(N-k+1)}{k}x>1$.

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
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