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Difference between revisions of "1992 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
  
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In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio <math>3: 4: 5</math>?
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In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
  
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== Solution ==
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<cmath>\begin{array}{c@{\hspace{8em}}
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c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}}
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c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}}
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c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt}
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\text{Row 0: } &    &    &    &    &    &    & 1 &    &    &    &    &    &  \\\vspace{4pt}
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\text{Row 1: } &    &    &    &    &    & 1 &    & 1  &    &    &    &    &  \\\vspace{4pt}
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\text{Row 2: } &    &    &    &    & 1 &    & 2 &    & 1  &    &    &    &  \\\vspace{4pt}
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\text{Row 3: } &    &    &    &  1 &    & 3 &    & 3  &    & 1 &    &    &  \\\vspace{4pt}
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\text{Row 4: } &    &    & 1  &    & 4 &    & 6 &    & 4  &    & 1 &    &  \\\vspace{4pt}
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\text{Row 5: } &    & 1 &    & 5  &    &10&    &10 &    & 5 &    & 1 &  \\\vspace{4pt}
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\text{Row 6: } & 1 &    & 6  &    &15&    &20&    &15 &    & 6 &    & 1
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\end{array}</cmath>
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In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio <math>3 :4 :5</math>?
  
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In Pascal's Triangle, we know that the binomial coefficients of the <math>n</math>th row are <math>\binom{n} {0}, \binom{n}{1}, ..., \binom{n} {n}</math>. Let our row be the <math>n</math>th row such that the three consecutive entries are <math>\binom{n} {r}</math>, <math>\binom{n}{r+1}</math> and <math>\binom{n} {r+2}</math>.  
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== Solution 1==
 +
 
 +
Consider what the ratio means. Since we know that they are consecutive terms, we can say
 +
<cmath>\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.</cmath>
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 +
Taking the first part, and using our expression for <math>n</math> choose <math>k</math>,
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<cmath> \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}</cmath>
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<cmath> \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} </cmath>
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<cmath> \frac{1}{3(n-k+1)} = \frac{1}{4k} </cmath>
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<cmath> n-k+1 = \frac{4k}{3} </cmath>
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<cmath> n = \frac{7k}{3} - 1 </cmath>
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<cmath> \frac{3(n+1)}{7} = k </cmath>
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Then, we can use the second part of the equation.
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<cmath> \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} </cmath>
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<cmath> \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} </cmath>
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<cmath> \frac{1}{4(n-k)} = \frac{1}{5(k+1)} </cmath>
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<cmath> \frac{4(n-k)}{5} = k+1 </cmath>
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<cmath> \frac{4n}{5}-\frac{4k}{5} = k+1 </cmath>
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<cmath> \frac{4n}{5} = \frac{9k}{5} +1. </cmath>
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Since we know <math>k = \frac{3(n+1)}{7}</math> we can plug this in, giving us
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<cmath> \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 </cmath>
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<cmath> 4n = 9\left(\frac{3(n+1)}{7}\right)+5 </cmath>
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<cmath> 7(4n - 5) = 27n+27 </cmath>
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<cmath> 28n - 35 = 27n+27 </cmath>
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<cmath> n = 62 </cmath>
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We can also evaluate for <math>k</math>, and find that <math>k = \frac{3(62+1)}{7} = 27.</math> Since we want <math>n</math>, however, our final answer is <math>\boxed{062.}</math> ~<math>\LaTeX</math> by ciceronii
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 +
==Solution 2==
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Call the row <math>x=t+k</math>, and the position of the terms <math>t-1, t, t+1</math>. Call the middle term in the ratio <math>N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}</math>. The first term is <math>N \frac{t}{k+1}</math>, and the final term is <math>N \frac{k}{t+1}</math>. Because we have the ratio <math>3:4:5</math>,
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<math>\frac{t}{k+1} = \frac{3}{4}</math> and <math>\frac{k}{t+1} = \frac{5}{4}</math>.
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<math>4t = 3k+3</math> and <math>4k= 5t+5</math>
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<math>4t-3k=3</math>
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<math>5t-4k=-5</math>
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Solve the equations to get <math> t= 27, k=35</math> and <math>x = t+k = \boxed{062}</math>.
 +
 
 +
-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava
  
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After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\frac 34=\frac{r+1}{n-r}</math> and <math>\frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>n = \boxed{062}</math>.
 
  
 
{{AIME box|year=1992|num-b=3|num-a=5}}
 
{{AIME box|year=1992|num-b=3|num-a=5}}
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 13:13, 20 April 2024

Problem

In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.

\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$?

Solution 1

Consider what the ratio means. Since we know that they are consecutive terms, we can say \[\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.\]

Taking the first part, and using our expression for $n$ choose $k$, \[\frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}\] \[\frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!}\] \[\frac{1}{3(n-k+1)} = \frac{1}{4k}\] \[n-k+1 = \frac{4k}{3}\] \[n = \frac{7k}{3} - 1\] \[\frac{3(n+1)}{7} = k\] Then, we can use the second part of the equation. \[\frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4(n-k)} = \frac{1}{5(k+1)}\] \[\frac{4(n-k)}{5} = k+1\] \[\frac{4n}{5}-\frac{4k}{5} = k+1\] \[\frac{4n}{5} = \frac{9k}{5} +1.\] Since we know $k = \frac{3(n+1)}{7}$ we can plug this in, giving us \[\frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1\] \[4n = 9\left(\frac{3(n+1)}{7}\right)+5\] \[7(4n - 5) = 27n+27\] \[28n - 35 = 27n+27\] \[n = 62\] We can also evaluate for $k$, and find that $k = \frac{3(62+1)}{7} = 27.$ Since we want $n$, however, our final answer is $\boxed{062.}$ ~$\LaTeX$ by ciceronii

Solution 2

Call the row $x=t+k$, and the position of the terms $t-1, t, t+1$. Call the middle term in the ratio $N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}$. The first term is $N \frac{t}{k+1}$, and the final term is $N \frac{k}{t+1}$. Because we have the ratio $3:4:5$,

$\frac{t}{k+1} = \frac{3}{4}$ and $\frac{k}{t+1} = \frac{5}{4}$.

$4t = 3k+3$ and $4k= 5t+5$

$4t-3k=3$ $5t-4k=-5$

Solve the equations to get $t= 27, k=35$ and $x = t+k = \boxed{062}$.

-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava


1992 AIME (Problems • Answer Key • Resources)
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Problem 5
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