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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-1}</math>, <math>\binom{n}{r}</math> and <math>\binom{n} {r+1}</math>. (We consider <math>r-1, r, r+1</math> because it cancels stuff out easier)
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== Solution 1==
  
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Consider what <math>\binom{n}{r}</math> equals to in a more explicit form. It equals <math>\frac{n!}{(n-r)!r!}</math>.  
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Consider what the ratio means. Since we know that they are consecutive terms, we can say
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<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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Now consider what it means to have three consecutive entries occurring in the ratio <math>3:4:5</math>. It means that we will have <math>\frac{\binom{n}{r-1}}{3} = \frac{\binom{n}{r}}{4} = \frac{\binom{n}{r+1}}{5}</math>. Note that the order of the ratio does not matter, as ascending from one side of Pascal's triangle is equivalent to descending from the opposite side of Pascal's triangle. We can multiply by a LCM of <math>60</math> to further simplify the problem into <math>20\binom{n}{r-1} = 15\binom{n}{r} = 12\binom{n}{r+1}</math>
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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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Using the more explicit form of <math>\binom{n}{r}</math>, we see that this equivalence function collapses into <math>\frac{20*n!*r}{(n-r+1)(n-r)!r!} = \frac{15*n!}{(n-r)!r!} = \frac{12*n!*(n-r)}{(n-r)!r!(r+1)}</math> (all of which is given by plugging in <math>r-1, r, r+1</math> into <math>\frac{n!}{(n-r)!r!}</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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After canceling out the <math>n!</math> in the numerator and the <math>(n-r)!r!</math> in the denominator, we get <math>\frac{20r}{n-r+1} = 15 = \frac{12(n-r)}{r+1}</math>. Setting the first equation to <math>15</math> and the third equation to <math>15</math>, we get a system that is solvable. We have: <cmath>20r = 15n - 15r + 15 \Rightarrow 35r - 15n = 15</cmath> <cmath>15r + 15 = 12n - 12r \Rightarrow 27r - 12n = -15</cmath>
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<math>4t = 3k+3</math> and <math>4k= 5t+5</math>
  
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Solving these equations, we get that <math>r = 27</math> and <math>n = 62</math>. Our goal is to find which row of Pascal's triangle this ratio occurs, or in other words find what n is, which we conclude to be <math>\boxed{62}.</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>.
 +
 
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-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava
  
  

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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