In how many ways can $1000 \mathrm{Ft}$ be made using only 1, 2, and $5 \mathrm{Ft}$ coins?
Solution. Let's count the cases when we use one- and two-forint coins for change, but not five-forint coins.

Using only one-forint coins represents one possibility. Then, let's replace two one-forint coins with one two-forint coin, and continue this process up to 4, 6, etc. This, together with the initial case, gives $\left[\frac{1000}{2}\right]+1=501$ possibilities.

Now, let's include five-forint coins among the change coins. Let $n$ be the sum of the values of the one- and two-forint coins, and $k$ be the number of five-forint coins. Then $n=1000-5 k$, where $0 \leq k \leq 200$. We previously saw that when $k=0$ (i.e., when there is no five-forint coin), the number of ways to make change is $\left[\frac{1000}{2}\right]+1$. In general, it is $\left[\frac{1000-5 k}{2}\right]+1$.

Let's write down the number of possible ways to make change for each possible value of $k$:

| $k$ | $n=1000-5 k$ | the number of possible ways to make change: $\left.\frac{1000-5 k}{2}\right]+1$ |  |
| :---: | :---: | :---: | :---: |
| 200 | 0 | 1 |  |
| 199 | 5 | 3 |  |
| 198 | 10 | 6 |  |
| $\vdots$ | $\vdots$ | $\vdots$ |  |
| 1 | 995 | 498 |  |
| 0 | 1000 | 501 |  |

It can be seen that from $k=200$ to 1, if we add the neighboring terms in pairs, we get a 100-term arithmetic sequence, whose first term is $a_{1}=4$ and the common difference is $d=10$. If we add these 100 terms and then add the value corresponding to $k_{0}$, we get the number of possible ways to make change:

$$
\sum_{k=0}^{200}\left(\left[\frac{1000-5 k}{2}\right]+1\right)=S_{100}+501=49901+501=50401
$$

Thus, there are 50401 ways to make 1000 Ft using one-, two-, and five-forint coins.