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

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== Solution ==
 
== Solution ==
  
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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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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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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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Taking the first part, and using our expression for <math>n</math> choose <math>k</math>,
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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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<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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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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<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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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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<cmath> n = \frac{7k}{3} - 1 </cmath>
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<cmath> \frac{3(n+1)}{7} = k </cmath>
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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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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
  
  

Revision as of 18:48, 31 March 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

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


1992 AIME (Problems • Answer Key • Resources)
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