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

Latest revision as of 11:13, 21 May 2020

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 $k=\boxed{166}$ .

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.

$\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}$

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.

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

~qwertysri987

@@ Line 7: / Line 7: @@
 == Solution ==
-Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have
+=== Solution 1 ===
+Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have
 <math>
-\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .
+\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]\, .
 </math>
-Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)x}{j}>1</math> in each of the terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the integer <math>k_{}^{}</math> that we are looking for must satisfy <math>k=\Big\lceil\frac{(N+1)x}{1+x}\Big\rceil</math>, where <math>\lceil z_{}^{}\rceil</math> denotes the greatest integer smaller than or equal to <math>z_{}^{}</math>.
+Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)x}{j}>1</math> in each of the <math>\log\big[\;\big]</math> terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the integer <math>k_{}^{}</math> that we are looking for must satisfy <math>k=\Big\lfloor\frac{(N+1)x}{1+x}\Big\rfloor</math>, where <math>\lfloor z_{}^{}\rfloor</math> denotes the greatest integer less than or equal to <math>z_{}^{}</math>.
-In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k_{}^{}=166</math>.
+In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k=\boxed{166}</math>.
+===Solution 2===
+We know that once we have found the largest value of <math>k</math>, all values after <math>A_k</math> are less than <math>A_k</math>. Therefore, we are looking for the smallest possible value such that:
+<math>{\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}</math>
+Dividing by <math>{\frac{1}{5}}^k</math> gives:
+<math>{1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}</math>.
+We can express these binomial coefficients as factorials.
+<math>\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}</math>
+We note that the <math>1000!</math> can cancel. Also, <math>(1000-k)!=(1000-k)(1000-k-1)!</math>. Similarly, <math>(k+1)!=(k+1)k!</math>.
+Canceling these terms yields,
+<math>\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}</math>
+Cross multiplying gives:
+<math>5k+5>1000-k \Rightarrow k>165.8</math>
+Therefore, since this identity holds for all values of <math>k>165.8</math>, the largest possible value of <math>k</math> is <math>\boxed{166}</math>.
+===Solution 3===
+We know that <math>A_k</math> will increase as <math>k</math> increases until certain <math>k=m</math>, where <math> A_0 < A_1 < A_2 < \dots < A_{m-2} < A_{m-1} < A_m </math> and
+<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 numerator must be greater than the
+denominator or
+<cmath>1000-k+1>5k</cmath>
+<cmath>k<166.8</cmath>.
+We conclude that the answer is <math>\boxed{166}</math>.
+==Solution 4==
+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>)
+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
+<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>
+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.
+==Solution 5==
+Notice the relation between <math>A_m</math> and <math>A_{m+1}</math>. We have that: <math>A_{m+1}  = A_m \cdot \frac{1}{5} \cdot \frac{1000-m}{m+1}</math>. This is true because from <math>A_m</math> to <math>A_{m+1}</math> we have to multiply by <math>\frac{1}{5}=0.2</math> once,and then we must resolve the factorial issue. To do this, we must realize that
+<cmath> {1000 \choose m+1} = \frac{1000!}{(m+1)!(1000-m-1)!} = \frac{1000 \cdots (1001-m)(1000-m)}{(m+1)!}</cmath>
+and that
+<cmath> {1000 \choose m} = \frac{1000!}{m!(1000-m)!} = \frac{1000 \cdots (1001-m)}{m!}</cmath>
+<cmath>\Longrightarrow {1000 \choose m} \cdot \frac{1000-m}{m+1}</cmath>
+So now, we must find out for which <math>m</math> is <math>1000-m< 5 \cdot (m+1)</math> (when this happens the numerator is less than the denominator for the fraction <math>\frac{1000-m}{m+1}</math> so then we will have <math>A_m > A_{m+1}</math>) Then we find that for all <math>m > 165.833333</math>, the above inequality (<math>1000-m< 5 \cdot (m+1)</math>) holds, so then <math>m=165</math> so <math>k=m+1= \boxed{166}</math>.
+~qwertysri987
 == See also ==
 {{AIME box|year=1991|num-b=2|num-a=4}}
+{{MAA Notice}}

1991 AIME (Problems • Answer Key • Resources)
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