You may also like

problem icon

Leftovers

Weekly Problem 26 - 2008
If $n$ is a positive integer, how many different values for the remainder are obtained when $n^2$ is divided by $n+4$?

Harmonic Triangle

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

This is the start of the harmonic triangle:

\begin{array}{ccccccccccc} & & & & &\frac{1}{1} & & & & & \\ & & & & \frac{1}{2} & & \frac{1}{2} & & & & \\ & & & \frac{1}{3} & &\frac{1}{6} & & \frac{1}{3} & & & \\ & & \frac{1}{4} & &\frac{1}{12} & & \frac{1}{12} & & \frac{1}{4} & & \\ & \frac{1}{5} & & \frac{1}{20} & & \frac{1}{30} & & \frac{1}{20} & & \frac{1}{5} & \\ \frac{1}{6} & & \frac{1}{30} & & \frac{1}{60} & & \frac{1}{60} & & \frac{1}{30} & & \frac{1}{6}\\ & & & & & \ldots& & & & & \end{array}

Each fraction is equal to the sum of the two fractions below it.

Look at the triangle above and check that the rule really does work.

Can you work out the next two rows?

The $n$th row starts with the fraction $\frac{1}{n}$, so the first diagonal goes:

$\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$...


Take a look at the second diagonal:

$\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $\frac{1}{20}$...

Can you see a pattern?
What fraction will appear in the second position on the nth row?

Can you prove it?

What about the third and fourth diagonals?

 

Click here for a poster of this problem.