3A. A certain item costing 455 denars was paid for with 5 denar coins and 10 and 50 denar banknotes, with a total of 23 coins and banknotes given. The number of 5 denar coins is less than the number of 10 denar banknotes. How many 5 denar coins and 10 and 50 denar banknotes were used to pay for the item?
Solution. Let the number of 5 denar coins be $x$, and the numbers of 10 and 50 denar banknotes be $y$ and $z$ respectively.

For a chosen $z$, the smallest possible sum that can be paid with the available banknotes and coins is $50z + (23-z) \cdot 5 = 45z + 115$. If $z \geq 8$, this sum will exceed 455 denars, so $z \leq 7$.

Similarly, for a chosen $z$, the largest sum that can be paid is $50z + (23-z) \cdot 10$ or $40z + 230$. If $z \leq 5$, this sum will be less than 455 denars, so $z \geq 6$.

Let $z=7$. From the conditions of the problem, we get the system:

$$
\left\{\begin{array}{l}
5 x + 10 y = 105 \\
x + y = 16
\end{array}\right.
$$

whose solution $x=11, y=5$ does not satisfy the conditions of the problem, as it is known that $x < y$.

Let $z=6$. From the conditions of the problem, we get the system:

$$
\left\{\begin{array}{l}
5 x + 10 y = 155 \\
x + y = 17
\end{array}\right.
$$

whose solution is $x=3, y=14$. This solution satisfies all the conditions of the problem.

Therefore, the purchase was paid with 3 coins of 5 denars, 14 banknotes of 10 denars, and 6 banknotes of 50 denars.