10.5. Find all natural numbers $\boldsymbol{n}$ such that $\boldsymbol{n}$ is equal to the sum of three numbers, the first of which is the largest divisor of the number $\boldsymbol{n} \boldsymbol{- 1}$, different from $\boldsymbol{n}-1$, the second is the largest divisor of the number $\boldsymbol{n}-\mathbf{2}$, different from $\boldsymbol{n}-\mathbf{2}$, and the third is the largest divisor of the number $\boldsymbol{n}-\mathbf{3}$, different from $\boldsymbol{n}-\mathbf{3}$.
Answer: 58 and 66.

Solution. The maximum divisor of a number, different from the number itself, is equal to the number divided by its smallest prime divisor. 1) If $n$ is odd, then the smallest prime divisors of the numbers $n-1$ and $n-3$ are 2, and the smallest prime divisor of $n-2$ is denoted by $p$ - an odd prime number. Then $n=\frac{n-1}{2}+\frac{n-2}{p}+\frac{n-3}{2}=n-2+\frac{n-2}{p}$, from which $n=2 p+2$ is an even number - a contradiction.

2) If $n$ is even, the smallest prime divisor of $n-2$ is 2, and the smallest prime divisors of $n-1$ and $n-3$ are denoted by $p$ and $q$ respectively - different odd prime numbers. If they were the same, the difference between $n-1$ and $n-3$, which is 2, would be divisible by the odd $p=q$, which is impossible.

Then $n=\frac{n-1}{p}+\frac{n-2}{2}+\frac{n-3}{q}=\left(\frac{1}{2}+\frac{1}{p}+\frac{1}{q}\right) n-\left(1+\frac{1}{p}+\frac{3}{q}\right)\frac{1}{2}$. Then the smaller of these numbers is less than 4, that is, equal to 3, and the second is less than 6, that is, equal to 5. Let's first take $p=3, q=5$, then $n=\frac{31}{30} n-\frac{29}{15}$, so the possible answer is $n=58$. Verification: $58=\frac{57}{3}+\frac{56}{2}+\frac{55}{5}=19+28+11$. - the equality is satisfied and the numbers 19, 28, and 11 are indeed the maximum proper divisors of the numbers 57, 56, and 55.

Now let $p=5, q=3$, then $n=\frac{31}{30} n-\frac{11}{5}$, so the possible answer is $n=66$. Verification: $66=\frac{65}{5}+\frac{64}{2}+\frac{63}{3}=13+32+21$. - the equality is satisfied and the numbers 13, 32, and 21 are indeed the maximum proper divisors of the numbers 65, 64, and 63.

Grading criteria. It is noted that the maximum divisor of a number, different from the number itself, is equal to the number divided by its smallest prime divisor: 1 point. Cases 1) and 2) are considered: 1 point. The impossibility of an odd $n$ is shown: 1 point. For an even $n$, the difference between $p$ and $q$ is justified: 1 point. It is justified that $p=3, q=5$ or $p=5, q=3$: 2 points. Verification is done with a comment that "the numbers 19, 28, and 11 are indeed the maximum proper divisors of the numbers 57, 56, and 55": 1 point.