Can a cube be divided into 49 smaller cubes?
Let's choose the edge of the cube to be 6 units. We will divide the cube into $3 \times 6 \times 6$, $2 \times 6 \times 6$, and $1 \times 6 \times 6$ rectangular prisms. Then, we will divide the $3 \times 6 \times 6$ prism into 4 pieces of $3 \times 3 \times 3$, the $2 \times 6 \times 6$ prism into 9 pieces of $2 \times 2 \times 2$, and the $1 \times 6 \times 6$ prism into 36 unit cubes. Thus, we have divided the original cube into exactly $4 + 9 + 36 = 49$ smaller cubes.

Remarks. 1. Zoltán Németh (Bp. Fazekas M. Főv. Gyak. Gimn., II. o. t.) dealt with the generalization of the problem and showed that for $n \geq 73$, a cube can be divided into $n$ smaller cubes. This can still be sharpened, as the statement is true for $n \geq 52$ as well. We only provide a sketch of the proof. First, we note that a cube can be divided into a) 8, b) 20, c) 38 smaller cubes. For this, we choose the edge of the cube to be one unit. In case a), we divide the cube into all $\frac{1}{2}$ edge-length cubes. In case b), we first place a $\frac{2}{3}$ edge-length cube inside the original cube such that one vertex of the two cubes coincides, then we fill the remaining part with $3^{3}-2^{3}=19$ $\frac{1}{3}$ edge-length cubes. In case c), we proceed similarly, but now we place a $\frac{3}{4}$ edge-length cube and $4^{3}-3^{3}=37$ $\frac{1}{4}$ edge-length cubes inside our unit cube. After this, our statement can be easily proven by repeatedly applying the division methods used in the problem and the three methods mentioned above; we leave this to the reader.

2. It is not known whether a cube can be divided into 51 smaller cubes.