51. An equation containing sums. Find $n$, if

$$
\frac{1^{3}+3^{3}+5^{3}+\cdots+(2 n-1)^{3}}{2^{3}+4^{3}+6^{3}+\cdots+(2 n)^{3}}=\frac{199}{242}
$$
51. First, note that if \(a: b = c: d\), then obviously \((a+b): b = (c+d): d\). Applying such a transformation to the fractions in our equation, we get

\[
\frac{1^{3}+2^{3}+3^{3}+\cdots+(2 n)^{3}}{2^{3}\left(1^{3}+2^{3}+3^{3}+\cdots+n^{3}\right)}=\frac{441}{242}
\]

Then, applying the known formula for the sum of cubes of natural numbers \({ }^{*}\), we get

\[
\begin{aligned}
\frac{(2 n)^{2}(2 n+1)^{2}}{4}: \frac{8 n^{2}(n+1)^{2}}{4} & =\frac{441}{242}=\frac{(21)^{2}}{2 \cdot(11)^{2}} \\
\frac{(2 n+1)^{2}}{(n+1)^{2}} & =\frac{(21)^{2}}{(11)^{2}}
\end{aligned}
\]

From this, it is clear that \(n=10\). (The negative square root leads to a negative \(n\).)

52*. If we place a stack of \((k-1)\) rectangular parallelepipeds, each of length \(2x\), on the \(k\)-th horizontal parallelepiped of the same kind and shift our stack until its center of gravity is vertically aligned with the edge of this \(k\)-th parallelepiped, then the center of gravity of the entire configuration will be horizontally distant from the edge of the \(k\)-th parallelepiped by a distance of \(\frac{x}{k}\).

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-091.jpg?height=205&width=534&top_left_y=837&top_left_x=767)

From this, it follows that \(n\) domino tiles in our stack (with \(x = 1\) inch) can be shifted between two vertical parallel planes, which are 1 inch apart, such that the total length of the "protruding" edges, moving from top to bottom, is expressed as \(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n-1}\). In this case, the domino tiles form a semi-arch that will be in equilibrium. This sum represents a partial sum of the harmonic series; since this series diverges, we can, by choosing a sufficiently large \(n\), achieve any horizontal distance between the edge of the top tile and the edge of the bottom tile.

To have the entire top tile extend beyond the edge of the bottom tile, \(n\) must be at least 5. In this case, the total length of the "protruding" edges will be \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\), or approximately 2.083 inches.

If we now shift our tiles so that all the \(2 \times 0.25\) faces are parallel to the same plane, and the center of gravity of each upper part of our stack is located above the corner of the nearest lower tile, then the horizontal distance between the corners of the topmost and bottommost tiles will increase compared to the previous case by \(\left[1^{2} + \left(\frac{1}{2}\right)^{2}\right]^{\frac{1}{2}} = \frac{\sqrt{5}}{2}\) times.

\[
\text { [P. M. E. J., 1, } 411 \text { (April 1954).] }
\]