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

m (Solution 2: fixed typo - min not max)
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We conclude that the answer is <math>\boxed{166}</math>.
 
We conclude that the answer is <math>\boxed{166}</math>.
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==Solution 4==
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Notice that the expansion is the largest the moment BEFORE the <math>nC_p < 5</math> (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 <math>nC_p*\frac{1}{5} > 1</math>)
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Say we have <math>{1000 \choose 5}</math>... this equals <math>\frac{1000*999*998*997*996}{5*4*3*2*1}</math>, and if we have k+1, then it's just going to be equivalent to multiplying this fraction by <math>\frac{995}{6}</math>. Notice that this fraction's numerator plus denominator is equal to <math>1001</math>. Calling the numerator x and the denominator y, we get that
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<cmath>x+y = 1001</cmath> <cmath>\frac{x}{y} > 5</cmath> <cmath>x>5y</cmath> <cmath>6y<1001</cmath> <cmath>y<166.83333</cmath> and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = <math>\boxed{166}</math>
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Note: It may be more intuitive to equate <math>x=5y</math> and then use test points to figure out which side of the inequality to lie on.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=2|num-a=4}}
 
{{AIME box|year=1991|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:47, 15 July 2019

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 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 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.

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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