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

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(Solution 2)
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===Solution 2===
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==Solution 2==
Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is <math>\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>  
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Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is <math>\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 3:4:5 ratio, <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 = 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 12:11, 24 August 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. This is $\dbinom{x}{t}$. 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 3:4:5 ratio, $\frac{x-t}{t+1} = \frac{4}{3}$ and $\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}$. Solve the equation to get get $t= 26$ and $x = 62$.

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions

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