10.5. Let $a, b, c$ and $d$ be natural numbers and $\frac{a d-1}{a+1}+\frac{b d-1}{b+1}+\frac{c d-1}{c+1}=d$. Find all values that the number $d$ can take.
Solution: We will transform the equation in an equivalent manner:

\[
\begin{gathered}
\frac{a d-1}{a+1}+\frac{b d-1}{b+1}+\frac{c d-1}{c+1}=d \\
d-\frac{d+1}{a+1}+d-\frac{d+1}{b+1}+d-\frac{d+1}{c+1}=d \\
2 d=(d+1)\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right) \\
\frac{2 d}{d+1}=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \\
2-\frac{2}{d+1}=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \\
\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{2}{d+1}=2
\end{gathered}
\]

Since all variables are natural numbers, each of the first three fractions on the left side does not exceed \(1 / 2\). Therefore, for the equality to hold, the condition \(\frac{2}{d+1} \geqslant 2-3 \cdot \frac{1}{2}=\frac{1}{2}\) must be satisfied, from which \(d \leqslant 3\), i.e., \(d \in\{1,2,3\}\). We will show that all such values of \(d\) are possible.

- \(d=3\) is achieved when \(a=b=c=1\). (There are no other options, which can be proven but is not required.)

- \(d=2\) is achieved, for example, when \(a=b=1\) and \(c=2\). (There are two other triples obtained by permuting the values of the unknowns.)

- \(d=1\) occurs in several cases. For example, when \(c=6, b=2, a=1\) or when \(b=c=3, a=1\). In total, there are 10 such cases.

Answer: \(d \in\{1,2,3\}\).

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Proven that \(d \leqslant 3\), but not justified that all values from the set \(\{1,2,3\}\) are possible | \(3-5\) points depending on the number of justified values |
| Justified all possible values for \(d\), but not proven that there are no others | 3 points |
| Justified some (not all) possible values for \(d\) | 2 points |
| Correct answer without justification | 1 point |
| Proof that \(d \leqslant 3\) is missing, but there are significant advances in this direction | +1 point |