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

Revision as of 02:55, 3 January 2016

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:

$5k+5>1000-k \Rightarrow k>165.8$

Therefore, since this identity holds for all values of $k>165.8$ , the largest possible value of $k$ is $\boxed{166}$ .

Solution 3

We know that $A_k$ will increase as $k$ increases until certain $k=m$ , where $A_0 < A_1 < A_2 < \dots < A_{m-2} < A_{m-1} < A_m$ and

$A_m > A_{m+1} > A_{m+2} > \dots > A_{1000}.$

Next, to change $A_{k-1}$ to $A_k$ , we multiply $A_{k-1}$ by $\frac{1000-k+1}{5k}$ . It follows that the numerator must be greater than the

denominator or

$1000-k+1>5k$ $k<166.8$ .

We conclude that the answer is $\boxed{166}$ .

@@ Line 50: / Line 50: @@
 <math>A_m > A_{m+1} > A_{m+2} > \dots > A_{1000}.</math>
-Next, to change <math> A_{k-1} </math> to <math> A_k </math>, we multiply <math> A_{k-1} </math> by <math>\frac{1000-k+1}{5k}</math>. It follows that the nominator must be greater than the
+Next, to change <math> A_{k-1} </math> to <math> A_k </math>, we multiply <math> A_{k-1} </math> by <math>\frac{1000-k+1}{5k}</math>. It follows that the numerator must be greater than the
 denominator or

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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