4[System of Numeration (other) $)$

Auto: Pooodiissiyy.k.

The Director of the Mint wants to issue coins of 12 denominations (each - a natural number of rubles) so that any amount from 1 to 6543 rubles can be paid without change, using no more than 8 coins. Can he do this?

(When paying an amount, several coins of the same denomination can be used.)
Note that $9^{4}=6561>6543$.

If we release coins of three denominations $-1, 3$, and 4 rubles, it is easy to check that any amount from 1 to 8 rubles can be paid without change using no more than two coins.

Suppose the Mint produces coins with denominations $9^{k}, 3 \cdot 9^{k}$, and $4 \cdot 9^{k}$ rubles for $k=0,1,2,3$. Any number $N$ from 1 to 6560 can be uniquely represented in the form $N=a_{3} \cdot 9^{3}+a_{2} \cdot 9^{2}+a_{1} \cdot 9+a_{0}$, where the numbers $a_{k}$ can take values from 0 to 8. As shown above, the sum $a_{k} \cdot 9^{k}$ can be obtained using no more than two coins. Thus, the entire sum $N$ can be obtained using no more than $4 \cdot 2=8$ coins of the specified denominations.

## Answer

It can.

Author: Fedulkin L.E.

A jeweler made 6 identical-looking silver ornaments weighing 22 g, 23 g, 24 g, 32 g, 34 g, and 36 g, and instructed his apprentice to engrave the weight on each ornament. Can the jeweler determine in two weighings on a balance scale without weights whether the apprentice has mixed up the ornaments?

## Solution

First weighing: Place the ornaments with the inscriptions 22 g, 23 g, and 24 g on one pan, and the ornaments with the inscriptions 34 g and 36 g on the other. The second pan will only outweigh the first if the groups of ornaments are correctly identified. In this case, the remaining ornament has a weight of 32 g.

Second weighing: Place the ornaments with the presumed weights of 24 g and 32 g (which we know for sure) on the first pan, and the ornaments with the presumed weights of 22 g and 34 g on the second pan. In this case, the first pan will weigh either 56 g (if everything is correct) or less (if the apprentice made a mistake), and the second pan will weigh either 56 g or more. Only if the weights of these four ornaments are correctly indicated will the balance be in equilibrium. In this case, the weights of the remaining ornaments are 23 g and 36 g.

## Answer

It can.