1991 AIME Problems/Problem 3

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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 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 $k=\boxed{166}$.

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 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.


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,


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 $\boxed{166}$.

Solution 4

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 5

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 $k=m+1= \boxed{166}$.


See also

1991 AIME (ProblemsAnswer KeyResources)
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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