10.5. Find the largest natural number $n$, for which the product of the numbers $n, n+1, n+2, \ldots, n+20$ is divisible by the square of one of them.
Answer: 20!.

Solution. For $n=20$ ! we have $\frac{n(n+1)(n+2) \ldots(n+20)}{n^{2}}=$ $=\frac{(n+1)(n+2) \ldots(n+20)}{20!}=C_{n+20}^{20}-$ an integer.

Now let $n>20$ ! and suppose $P=n(n+1)(n+2) \ldots(n+20)$ is divisible by $k^{2}$, where $k=n+i, i \in\{0,1,2, \ldots, 20\}$. We have $P / k=$ $=(k-i)(k-i+1) \ldots(k-1)(k+1)(k+2) \ldots(k+j)$, where $j=20-i$. Note that the number $P / k \equiv(-1)^{i} i!j!(\bmod k)$ must be divisible by $k$. But $0<i!j!\leqslant i!\cdot(i+1)(i+2) \ldots(i+j)=20!<n \leqslant k$, so $i!j!$ does not divide $k$. Contradiction.

Comment. Only (correct) answer -1 point.

Verification that the answer fits -1 point.

Evaluation -5 points.

Unperformed calculations requiring significant work by the jury to check -1 point deducted.

Arithmetic error leading to incorrect calculation of one of the exponents -2 points deducted.

The inequality $k!(20-k)!\leqslant 20$ ! is accepted without proof if clearly stated.

Using but not proving that the product of $|i-k|$ over all $i$, except $k$, is less than $20!$ -1 point deducted.

Claiming that if $(x, y)=d$, then $x / d$ is coprime with $y$ -2 points deducted.

Incorrect work with inequalities when calculating the power of entry -1 point deducted.