AIME 1991 · 第 3 题

Solutions

Solution 1 (Proof Format)

$\textbf{Lemma:}$

The "maxima" is the point $k$ in which every $n > k$ results in a decrease of value.

$\textbf{Conjecture:}$

For all values $k$ upon $A_k$ there exists a value $n=k$ in which

$$\binom{1000}{n} \cdot \left(\frac{1}{5}\right)^n$$ is as great as possible.

$\textbf{Proof of the Conjecture}$

For simplicity, denote $a_k = \binom{1000}{k} \cdot \left(\frac{1}{5}\right)^k$.

Examine the ratio $\frac{a_{k+1}}{a_k}$. We see this simplifies into

$$\frac{a_{k+1}}{a_k} = \frac{\binom{1000}{k+1}\left(\frac{1}{5}\right)^{k+1}}{\binom{1000}{k}\left(\frac{1}{5}\right)^k} = \frac{\binom{1000}{k+1}}{\binom{1000}{k}} \cdot \frac{1}{5}$$ Then, the identity $\frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n-k}{k+1}$ simplifies the above ratio as

$$\frac{a_{k+1}}{a_k} = \frac{1000-k}{k+1} \cdot \frac{1}{5}$$ which can be rewritten as

$$\frac{1000-k}{5(k+1)}$$ Note that $5(k+1)$ will exponentially increase greater than $1000-k$ through properties of multiplication.

So, for the sequence to begin descent, we want the exact value of $k$ in which

$1000-k \le 5(k+1)$

Solving for which we obtain

$k \ge 165 \frac{5}{6}$.

So there exists some $n=k$ for which

$\binom{1000}{n} \cdot \left(\frac{1}{5}\right)^n$

is as great as possible.

particularly $n \ge 165 \frac{5}{6}$

$\textbf{Incorporation of the Conjecture}$

Every term in the binomial expansion of $(1+0.2)^{1000}$ must be in the form

$$A_k = \binom{1000}{k}(0.2)^k$$ where $k \in \mathbb{Z}^+ : 0 \le k \le 1000$.

The maximal value occurs at $k \ge 165 \frac{5}{6}$. However, this isn't an integer. The next greatest integer will occur at $k = 166$. Thus the value of $k$ such that $A_k$ is maximized is $\boxed{\textbf{166}}$ $\square$

~Pinotation

Solution 2

Let $0. Then we may write$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 $0). Therefore, the integer$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 $k=\boxed{166}$.

Solution 3

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 smallest 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 4

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 $\boxed{166}$.

Solution 5

Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\frac{1}{5} > 1$)

Say we have ${1000 \choose 5}$... this equals $\frac{1000*999*998*997*996}{5*4*3*2*1}$, and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\frac{995}{6}$. Notice that this fraction's numerator plus denominator is equal to $1001$. Calling the numerator x and the denominator y, we get that

$$x+y = 1001$$ $$\frac{x}{y} > 5$$ $$x>5y$$ $$6y<1001$$ $$y<166.83333$$ and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\boxed{166}$

Note: It may be more intuitive to equate $x=5y$ and then use test points to figure out which side of the inequality to lie on.

Solution 6

Notice the relation between $A_m$ and $A_{m+1}$. We have that: $A_{m+1} = A_m \cdot \frac{1}{5} \cdot \frac{1000-m}{m+1}$. This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that

$${1000 \choose m+1} = \frac{1000!}{(m+1)!(1000-m-1)!} = \frac{1000 \cdots (1001-m)(1000-m)}{(m+1)!}$$ and that

$${1000 \choose m} = \frac{1000!}{m!(1000-m)!} = \frac{1000 \cdots (1001-m)}{m!}$$ $$\Longrightarrow {1000 \choose m} \cdot \frac{1000-m}{m+1}$$ So now, we must find out for which $m$ is $1000-m< 5 \cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$) Then we find that for all $m > 165.833333$, the above inequality ($1000-m< 5 \cdot (m+1)$) holds, so then $m=165$ so $k=m+1= \boxed{166}$.

~qwertysri987