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

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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>?
 
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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== Solution ==
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== Solution 1==
  
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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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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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==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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<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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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
  
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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{62}</math>.
 
  
 
{{AIME box|year=1992|num-b=3|num-a=5}}
 
{{AIME box|year=1992|num-b=3|num-a=5}}

Revision as of 14:32, 22 November 2020

Problem

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 $3: 4: 5$?

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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