1992 AIME Problems/Problem 4

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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