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The Head of the Mint wants to issue coins of 12 denominations (each in 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 will be possible.