109 individuals purchased 109 books for a total value of $2845 \mathrm{Ft}$ to give as gifts. It turned out that only three different prices appeared among the books: $34 \mathrm{Ft}, 27.50 \mathrm{Ft}$, and $17.50 \mathrm{Ft}$. Determine how many books belonged to the first, second, and third price category, knowing that the number of books with the same price did not differ much from each other.
I. Solution. Let the number of books priced at 34 Ft be $x$, the number of books priced at 17.50 Ft be $y$, then the number of books priced at 27.50 Ft is $109-x-y$, and according to the bill, after rearrangement,

$$
\begin{gathered}
34 x+17.5 y+27.5(109-x-y)=2845 \\
6.5 x-10 y=-152.5 \\
x=\frac{20 y-305}{13}=y-23+\frac{7 y-6}{13}
\end{gathered}
$$

$x$ and $y$ are natural numbers, so the third term on the right side must be a natural number $u$:

$$
\frac{7 y-6}{13}=u, \text { from which } y=\frac{13 u+6}{7}=2 u+1-\frac{u+1}{7}
$$

and repeating this thought process,

$$
\begin{aligned}
& \frac{u+1}{7}=v, \text { an integer, from which } u=7 v-1, \text { and furthermore } \\
& \qquad \begin{array}{l}
y=2 u+1-v=13 v-1 \\
x=y-23+u=20 v-25
\end{array}
\end{aligned}
$$

finally, the number of books priced at 27.50 Ft is $z=135-33 v$.

We only get a positive integer for $x$ if $v$ is at least 2; however, $z$ remains positive only as long as $v$ is not greater than 4, so only 2, 3, and 4 can be used for $v$ and

$$
\begin{aligned}
& v=2 \text { in which case } \\
& x=15 \\
& y=25 \\
& z=69 \text {, } \quad \\
& v=3 \quad " \quad x=35 \\
& y=38 \\
& z=36 \\
& v=4 \quad " \\
& x=55 \\
& y=51 \\
& z=3
\end{aligned}
$$

Only the set of values derived from $v=3$ meets the final requirement of the problem, according to which they bought 35 books priced at 34 Ft, 38 books priced at 27.50 Ft, and 36 books priced at 17.50 Ft.

János Bárdos (Makó, József A. g. I. o. t.)

II. Solution. Using the same notation as in the first solution and a similar thought process, we arrive at the following equation:

$$
33 x+20 z=1875
$$

The first term on the left side is divisible by 3, the second term by 5, and the right side by both. Since 3 and 5 are relatively prime numbers, this is only possible if $x$ is a multiple of 5 and $z$ is a multiple of 3: $x=5 t, z=3 s$, where $t$ and $s$ are natural numbers. Substituting these into (1) and simplifying,

$$
11 t+4 s=125
$$

The right side is odd, but $4 s$ is even, so $11t$, and thus $t$ itself, must be odd: $t=2 r-1$, where $r$ is a natural number. Substituting this into (2) and rearranging,

$$
11 r+2 s=68
$$

Here, however, $r$ can only be even, $r=2 q$ ( $q$ is a natural number), from which $s=34-11 q$.

Accordingly, only $q=1,2,3$ meet the requirement, from which $r=2,4,6 ; t=3,7,11 ; x=15,35,55$; and $s=23,12,1 ; z=69,36,3$. From here, using the thought process of the first solution, we arrive at the result.

III. Solution. Start from the last statement of the problem. The average price of the books is $(34+27.50+17.50) / 3=79 / 3$, which is almost exactly 36th of the spent amount, so by taking 36 of each type of book, we would spend $36(34+27.5+17.5)=2844 \mathrm{Ft}$, leaving us with 1 Ft and we would still need to buy one more book. We need to ensure at least 16.50 Ft for the cover of the last book, which is only possible by exchanging a more expensive book for a cheaper one. The difference between the highest and lowest unit price is exactly $16.50 \mathrm{Ft}$, so instead of 1 book priced at 34 Ft, we can buy 2 books priced at 17.50 Ft, using the 1 Ft remainder, the money runs out, and we have the desired 109 books.

We cannot get a solution with fewer exchanges, so according to the prices in descending order, they bought 35, 36, and 38 books.

Ottó Dobozi (Budapest, Apáczai Csere J. gyak. g. I. o. t.)

Note. Many papers stated that for the books priced at 27.50 Ft and 17.50 Ft, which include cents, we must buy an even number of each to use the whole amount of forints. - This is incorrect. It is true that $y+z$ is even and $x$ is odd, but this does not make the solution unique.