We have bricks at our disposal, each with a length of $22 \mathrm{~cm}$, a width of $11 \mathrm{~cm}$, and a height of $6 \mathrm{~cm}$. We need to fill a rectangular parallelepiped - $A B C D E F G H$ - with these bricks such that the $22 \mathrm{~cm}$ edges of the bricks are parallel to the length $A B$ of the stack, the $11 \mathrm{~cm}$ edges are parallel to the width $B F$ of the stack, and the $6 \mathrm{~cm}$ edges are parallel to the height $A D$ of the stack. The length $A B$ is $1 \frac{1}{4}$ times the height $A D$, and $1 \frac{1}{2}$ times the width $B F$. What is the smallest number of bricks that meets the above requirements?
Let the length of $AB$ be $a$, the width of $BF$ be $b$, and the height of $AD$ be $c$. According to our data,

$$
a=\frac{3}{2} b=\frac{5}{4} c \quad \text { or } \quad 4 a=6 b=5 c
$$

Suppose we need to place $n$ bricks side by side in the $AB$ direction, $p$ bricks in the $BF$ direction, and $q$ bricks in the $AD$ direction; then

$$
a=22 n, \quad b=11 p, \quad c=6 q
$$

It is necessary that

$$
4 \cdot 22 \cdot n=6 \cdot 11 \cdot p=5 \cdot 6 \cdot q
$$

Since $6 \cdot 11 \cdot p=5 \cdot 6 \cdot q$, or $11 p=5 q$, it is necessary and sufficient that $p=5 k, q=11 k$, where $k$ is an ordinary integer.

For $8 \cdot 11 \cdot n=6 \cdot 11 \cdot p$, or $4 n=3 p$, it is necessary that $n=3 l, p=4 l$, where $l$ is an ordinary integer. Finally, equating the values of $p$, it is necessary that

$$
5 k=4 l, \quad \text { i.e., } k=4 m \text { and } l=5 m \text { should be. }
$$

According to these, the number of bricks needed to fill the parallelepiped-shaped stack in the three directions is:

$$
n=3 \cdot 5 \cdot m, \quad p=4 \cdot 5 \cdot m, \quad q=4 \cdot 11 \cdot m
$$

where $m$ is an ordinary integer. The smallest values of $n, p, q$ occur when $m=1$; these are

$$
n=15, \quad p=20, \quad q=44
$$

The number of bricks: $N=15 \cdot 20 \cdot 44=13200$ and the dimensions of the stack are:

$$
a=15 \cdot 22 \text{ cm}=3.3 \text{ m}, \quad b=20 \cdot 11 \text{ cm}=2.2 \text{ m}, \quad c=44 \cdot 6 \text{ cm}=264 \text{ cm}
$$

Béla Gerő and Ödön Schläffer (Bencés g. VIII. o. Esztergom.)