1992 AIME Problems/Problem 4

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 $N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$

 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