# 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 (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions