Dwarves count their ages in days, so they celebrate their birthdays every day. At Nosey the Dwarf's party, seven dwarves with ages of 105, 120, 140, 168, 210, 280, and 420 days gathered. During the celebration, all eight dwarves managed to divide themselves into two groups with equal sums of ages.

What is the minimum and maximum number of days Nosey the Dwarf could have been?

(E. Novotná)
The sum of the ages of the seven dwarfs who came to visit Pinocchio is 1443 days. If the dwarfs were divided so that one group consisted of Pinocchio alone and the other group consisted of all the others, Pinocchio would be 1443 days old. Pinocchio cannot be older than 1443 days.

The sum of the ages of the seven dwarfs who came to visit Pinocchio is an odd number of days. For the sum of the ages of all eight dwarfs to be divisible by two, Pinocchio's number of days must also be odd. To be able to divide the dwarfs into two groups with the same ages, the sum of the seven known ages must be able to express both half the total sum and half of this sum without Pinocchio's age. We will sequentially examine the possibilities until we find a suitable one. In the following table, $N$ represents Pinocchio's age and $S$ the sum of the ages of all eight dwarfs. The decision in the last row is explained below:

| $N$ | 1 | 3 | 5 | 7 | 9 | 11 | 13 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $S$ | 1444 | 1446 | 1448 | 1450 | 1452 | 1454 | 1456 | $\ldots$ |
| $\frac{1}{2} S$ | 722 | 723 | 724 | 725 | 726 | 727 | 728 | $\ldots$ |
| $\frac{1}{2} S-N$ | 721 | 720 | 719 | 718 | 717 | 716 | 715 | $\ldots$ |
| $?$ | $\check{c}$ | $\mathrm{~s}$ | $\check{c}$ | $\mathrm{~s}$ | $\check{\mathrm{c}}$ | č | $\mathrm{OK}$ | $\ldots$ |

Apart from the two dwarfs who came to visit Pinocchio, the ages of all others are divisible by 10. The ages of these two exceptions end in the digits 5 and 8. Therefore, from the seven known ages of the dwarfs, only sums ending in the digits 0, 5, 8, and 3 (depending on whether none, one, or both of the mentioned numbers are included) can be expressed. This observation rules out the possibilities corresponding to $N=1, 5, 9$, 11 (and many others). We will examine the remaining possibilities:

- In the case of $N=3$, the sum $\frac{1}{2} S=723$ must include both numbers 105 and 168:

$$
723=105+168+450
$$

The summand 450 must be expressible using the numbers 120, 140, 210, 280, and 420. By sequential testing (e.g., starting from the largest available number), we find that this is not possible.

- In the case of $N=7$, the sum $\frac{1}{2} S-N=718$ must include the number 168:

$$
718=168+550 .
$$

The summand 550 must be expressible using the numbers 120, 140, 210, 280, and 420. By sequential testing (e.g., starting from the largest available number), we find that this is not possible.

- In the case of $N=13$, the sum $\frac{1}{2} S=728$ must include the number 168:

$$
728=168+560 \text {. }
$$

The summand 560 must be expressible using the numbers 120, 140, 210, 280, and 420. By testing, we find $560=140+420$, so we have found the smallest suitable solution.

Pinocchio could be at least 13 days old and at most 1443 days old.

Notes. The division into groups with a 13-day-old Pinocchio (i.e., the division corresponding to the sum $\left.\frac{1}{2} S=728\right)$ looked like this:

$$
140+168+420=13+105+120+210+280 .
$$

In the cases of $N=3$ and 7, it is possible to focus on the second discussed sum (i.e., 720 and 725, respectively), which also cannot be expressed using the available numbers.

In the process of eliminating possibilities, other properties of the given numbers can be used. For example, the impossibility of expressing the number 450 in the case of $N=3$ can be justified as follows. The number 450 is divisible by 10 but not by 20. Of the available numbers, only the number 210 has the same property, so it must be included in the expression of 450:

$$
450=210+240 .
$$

The summand 240 must be expressible using the remaining numbers 120, 140, 280, and 420. Since the numbers 280 and 420 are greater than 240, only the sum $120+140$ is possible. However, this sum is also greater than 240, so the expression of 450 is not possible.