## 3. Hussar

The captain of a hussar ship divides gold coins with his thirteen hussars in the following way:

first division: one for me, one for each of you second division: two for me, one for each of you third division: three for me, one for each of you.

In each subsequent division, the captain takes one more gold coin than in the previous division, and gives each of the other hussars one gold coin.

How many divisions are needed for the captain to have exactly 187500 more gold coins than all his hussars combined?

## Result: $\quad 625$
## Solution.

In $n$ divisions, the captain receives $1+2+3+4+\cdots+n$ gold coins, and each pirate gets $n$ gold coins.

We need to determine the number $n$ for which $1+2+3+4+\cdots+n-13n=187500$ holds.

Using Gauss's insight, we get:

$$
\frac{n \cdot(n+1)}{2}-13n=187500
$$

which simplifies to

$$
\begin{gathered}
n \cdot(n+1)-26n=375000 \\
n \cdot(n-25)=375000
\end{gathered}
$$

We need to represent the number 375000 as the product of two natural numbers that differ by 25.

Factorize 375000: $375000=2^{3} \cdot 3 \cdot 5^{6}=25 \cdot 25 \cdot\left(5^{2} \cdot 2^{3} \cdot 3\right)$.

We see that $375000=625 \cdot 600$, so the solution to the equation is $n=625$.

It takes 625 divisions for the captain to have 187500 more gold coins than all his pirates combined.