AoPSWiki
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

1991 AIME Problems/Problem 3

From AoPSWiki

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}^{}={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\l...

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.

See also

1991 AIME (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_3"
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us