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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> ~LaTeX 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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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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==Solution 2==
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Call the row x, and the number from the leftmost side t. Call the first term in the ratio <math>N</math>. <math>N = \dbinom{x}{t}</math>. The next term is <math>N * \frac{x-t}{t+1}</math>, and the final term is <math>N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math>. Because we have the ratio, <math>N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math> = <math>3:4:5</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>\frac{x-t}{t+1} = \frac{4}{3}</math> and <math>\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}</math>.
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Solve the equation to get get <math> t= 26 </math> and <math>x = \boxed{062}</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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https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3
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-Solution and LaTeX by jackshi2006
  
  

Revision as of 00:11, 1 August 2021

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, and the number from the leftmost side t. Call the first term in the ratio $N$. $N = \dbinom{x}{t}$. The next term is $N * \frac{x-t}{t+1}$, and the final term is $N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$. Because we have the ratio, $N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$ = $3:4:5$,

$\frac{x-t}{t+1} = \frac{4}{3}$ and $\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}$. Solve the equation to get get $t= 26$ and $x = \boxed{062}$.

https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3


-Solution and LaTeX by jackshi2006


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