Question

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.

$\begin{array}{rl}0:&1\\1:&1\quad1\\2:&1\quad2\quad1\\3:&1\quad3\quad3\quad1\\4:&1\quad4\quad6\quad4\quad1\\5:&1\quad5\quad10\quad10\quad5\quad1\\6:&1\quad6\quad15\quad20\quad15\quad6\quad1\end{array}$

In which row of Pascal’s Triangle do three consecutive entries occur that are in the ratio $3:4:5?$

Answer

Answer: 62

Concepts:Pascal’s Triangle combinations system of equations

Difficulty rating: 1940

Small Hint:

Represent the three entries as $\binom nk,$ $\binom n{k+1},$ and $\binom n{k+2}$

Big Hint:

Use the ratios of consecutive binomial coefficients to obtain two linear equations in $n$ and $k$

Solution:

For three consecutive entries beginning at position $k,$ $\begin{aligned}\frac{\binom n{k+1}}{\binom nk}&=\frac{n-k}{k+1}=\frac43,\\\frac{\binom n{k+2}}{\binom n{k+1}}&=\frac{n-k-1}{k+2}=\frac54.\end{aligned}$ Thus $3n=7k+4$ and $4n=9k+14.$ Solving gives $k=26$ and $n=62.$

1992 AIME Problem 4

Answer: 62

Difficulty rating: 1940

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years