The price of $1,2,3, \ldots, 9,10$ units of two foreign goods, rounded to the nearest whole forint, is as follows:

| quantity: | 1, | 2, | 3, | 4, | 5, | 6, | 7, | 8, | 9, | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Item I: | $\mathbf{4 6}$, | 91, | $\mathbf{1 3 7}$, | 182, | $\mathbf{2 2 8}$, | 273, | $\mathbf{3 1 9}$, | 364, | 410, | $\mathbf{4 5 6}$ |
| Item II: | 73, | $\mathbf{1 4 7}$, | 220, | $\mathbf{2 9 4}$, | 367, | $\mathbf{4 4 1}$, | 514, | $\mathbf{5 8 8}$, | 661, | $\mathbf{7 3 5}$. |

The prices written in bold digits were obtained by rounding up.

Based on this, how accurately can the unrounded prices of these two items be determined?
Let the unit price of the first item be $x$ Ft, and that of the second item be $y$ Ft. Each data point provides a lower and an upper limit for the price of the respective number of items, the difference of which, considering the direction of rounding, is $0.50 \mathrm{Ft}$, for example:

$$
227.5 \leq 5 x < 228.00, \quad 410.00 \leq 9 x \leq 410.50
$$

Dividing these by the number of items, we get a pair of limits for the unit price; the difference of which is already smaller:

$$
45.50 \leq x < 45.60, \quad 45.555 \ldots < x < 45.611 \ldots
$$

Based on these, we can write a single double inequality, where the lower limit is the greater of the two previous lower limits, and the upper limit is the smaller of the two previous upper limits:

$$
45.555 \ldots < x < 45.60
$$

Finding the greatest of the 10 lower limits (which comes from $9x$), and the smallest of the 10 upper limits (from $8x$), the limits of the unrounded price of the first item are:

$$
45 \frac{5}{9} = 45.555 \ldots \leq x \leq 45.5625 = 45 \frac{9}{16}
$$

Finally, the difference between the two limits is $1 / 144 (<0.007) \mathrm{Ft}$.

For the unit price of the second item, a similar calculation step by step yields a progressively larger lower limit, while the upper limit remains the same:

$$
73.45 = 73 \frac{9}{20} < y \leq 73 \frac{1}{2} = 73.50
$$

Here, the difference between the two limits is $1 / 20 = 0.05 \mathrm{Ft}$.

Gábor Dombi (Szeged, Ságvári E. Gymnasium, 2nd grade)