. A mason has bricks with dimensions $2 \times 5 \times 8$ and other bricks with dimensions $2 \times 3 \times 7$. She also has a box with dimensions $10 \times 11 \times 14$. The bricks and the box are all rectangular parallelepipeds. The mason wants to pack bricks into the box filling its entire volume and with no bricks sticking out. Find all possible values of the total number of bricks that she can pack.
Answer: 24.

Let the number of $2 \times 5 \times 8$ bricks in the box be $x$, and the number of $2 \times 3 \times 7$ bricks $y$. We must figure out the sum $x+y$. The volume of the box is divisible by 7 , and so is the volume of any $2 \times 3 \times 7$ brick. The volume of a $2 \times 5 \times 8$ brick is not divisible by 7 , which means that $x$ must be divisible by 7 .

The volume of the box is $10 \cdot 11 \cdot 14$. The volume of the $2 \times 5 \times 8$ bricks in the box is $x \cdot 2 \cdot 5 \cdot 8=80 x$. Since this volume cannot exceed the volume of the box, we must have

$$
x \leqslant \frac{10 \cdot 11 \cdot 14}{80}=\frac{11 \cdot 7}{4}=\frac{77}{4}<20 .
$$

Since $x$ was divisible by 7 , and certainly nonnegative, we conclude that $x$ must be 0,7 or 14 . Let us explore each of these possibilities separately.

If we had $x=0$, then the volume of the $2 \times 3 \times 7$ bricks, which is $y \cdot 2 \cdot 3 \cdot 7$, would be equal to the volume of the box, which is $10 \cdot 11 \cdot 14$. However, this is not possible since the volume of the $2 \times 3 \times 7$ bricks is divisible by three whereas the volume of the box is not. Thus $x$ must be 7 or 14 .

If we had $x=7$, then equating the total volume of the bricks with the volume of the box would give

$$
7 \cdot 2 \cdot 5 \cdot 8+y \cdot 2 \cdot 3 \cdot 7=10 \cdot 11 \cdot 14
$$

so that

$$
y \cdot 2 \cdot 3 \cdot 7=10 \cdot 11 \cdot 14-7 \cdot 2 \cdot 5 \cdot 8=1540-560=980 .
$$

However, again the left-hand side, the volume of the $2 \times 3 \times 7$ bricks, is divisible by three, whereas the right-hand side, 980, is not. Thus we cannot have have $x=7$ either, and the only possibility is $x=14$.

Since $x=14$, equating the volumes of the bricks and the box gives

$$
14 \cdot 2 \cdot 5 \cdot 8+y \cdot 2 \cdot 3 \cdot 7=10 \cdot 11 \cdot 14
$$

which in turn leads to

$$
y \cdot 2 \cdot 3 \cdot 7=10 \cdot 11 \cdot 14-14 \cdot 2 \cdot 5 \cdot 8=1540-1120=420
$$

so that

$$
y=\frac{420}{2 \cdot 3 \cdot 7}=\frac{420}{42}=10
$$

Thus the number of bricks in the box can only be $14+10=24$. Finally, for completeness, let us observe that 14 bricks with dimensions $2 \times 5 \times 8$ can be used to fill a volume with dimensions $10 \times 8 \times 14$, and 10 bricks with dimensions $2 \times 3 \times 7$ can be used to fill a volume with dimensions $10 \times 3 \times 14$, so that these 24 bricks can indeed be packed in the box.