11.6. Given a natural number $n$. Sasha claims that for any $n$ rays in space, no two of which have common points, he can mark $k$ points on these rays lying on the same sphere. For what largest $k$ is his statement true?

(A. Kuznetsov)
Answer. $k=n$ for even $n$, $k=n+1$ for odd $n$, that is, $2 \cdot\left\lceil\frac{n}{2}\right\rceil$.

Solution. Example. For even $n=2 m$, consider $m$ parallel lines and on each select a pair of non-intersecting rays. Note that in each pair of rays, there are no more than two intersections with the sphere, since a line has no more than two common points with a sphere, so $k \leqslant 2 m$. The example for odd $n=2 m-1$ is obtained by removing one ray from the example for $n=2 m$.

Estimate. Consider some line $\ell$ that is not perpendicular to any of our rays. Consider the projections of our rays onto $\ell$, among which at least $\left\lceil\frac{n}{2}\right\rceil=\left[\frac{n+1}{2}\right]$ are directed in one direction (we will say to the right), and forget about the other rays. Let point $X$ on the line belong to all the selected projections, choose an arbitrary point $Y \in \ell$ to the right. Let $\alpha_{X}$ and $\alpha_{Y}$ be planes perpendicular to the line $\ell$, passing through $X$ and $Y$ respectively. Each of the selected rays intersects both these planes. Choose a sufficiently large $R$ such that the circle $\omega \subset \alpha_{Y}$ with center $Y$ and radius $R$ contains all the intersection points of the plane $\alpha_{Y}$ with the selected rays. Consider the sphere $\Omega$ that is tangent to the plane $\alpha_{X}$ at point $X$ and contains the circle $\omega$. Consider any of our rays. It passes through a point inside the sphere $\Omega$, and its origin lies in a different half-space relative to the plane $\alpha_{X}$ than $\Omega$, so it intersects $\Omega$ at two points. Thus, we have obtained $k=2\left\lceil\frac{n}{2}\right\rceil$ intersection points. 11.7. For a natural number $N$, consider all different perfect squares that can be obtained from $N$ by erasing one digit in its decimal representation. Prove that the number of these squares does not exceed some value independent of $N$.

(S. Berlov, F. Petrov, D. Krachun)

Solution. Let the number $N$ consist of $k+1$ digits. We will assume that $k>100$; smaller numbers do not affect the desired boundedness.

For $i=1, \ldots, k$, let $n_{i}$ be the number obtained by removing the $i$-th digit from the end of $N$. Denote by $f(N)$ the number of perfect squares in the set $\left\{n_{1}, \ldots, n_{k}\right\}$. Our goal is to prove that $f(N)$ is bounded above.

Let $N=10^{t} N_{1}$, where $N_{1}$ is not divisible by 10. If $t$ is odd, then the number $n_{i}$ can be a perfect square only for $i \leqslant t+1$, so in this case $f(N) \leqslant 2$. If $t$ is even, then the final $t$ zeros do not affect the matter, so $f(N)=f\left(N_{1}\right)$. Therefore, we will assume that $N$ is not divisible by 10.

Let us select the set $A \subset\{1, \ldots, k\}$ of $f(N)$ indices $i$ for which $n_{i}=m_{i}^{2}$ is a perfect square, and the natural numbers $m_{i}, i \in A$, are pairwise distinct.

Note the following:

1) $n_{i} \geqslant 10^{k-1}$, hence $m_{i} \geqslant 10^{(k-1) / 2}$ for all $i \in A$;
2) $\left|n_{i}-n_{j}\right| \leqslant 10^{k-1}$, hence $\left|m_{i}^{2}-m_{j}^{2}\right| \leqslant 10^{k-1}$, and since $m_{i} \geqslant 10^{(k-1) / 2}$, we have $\left|m_{i}-m_{j}\right| \leqslant 10^{(k-1) / 2}$. Thus, all elements of $A$, except possibly one, are greater than $(k-1) / 2$. Denote $A_{1}:=A \backslash\{\min (A)\}$ (remove the smallest element from $A$), then $\left|A_{1}\right|=f(N)-1$ and $\min \left(A_{1}\right) \geqslant k / 2$.

Let $j>i$ be two elements of the set $A_{1}$. Then by properties 1) and 2) we have

$$
10^{j}>\left|n_{i}-n_{j}\right|=\left|m_{i}^{2}-m_{j}^{2}\right| \geqslant 2 \cdot 10^{(k-1) / 2} \cdot\left|m_{i}-m_{j}\right| .
$$

On the other hand, by property 3), the number $n_{i}-n_{j}=\left(m_{i}-m_{j}\right)\left(m_{i}+m_{j}\right)$ is divisible by $10^{i-1}$.

Let $r=\lceil(i-1) / 2\rceil$ (where $\lceil\cdot\rceil$ denotes the ceiling function). At least one of the numbers $m_{i}-m_{j}, m_{i}+m_{j}$ is divisible by $2^{r}$, and at least one is divisible by $5^{r}$. Moreover, if $N$ is odd, then the numbers $m_{i}, m_{j}$ are odd, so one of the numbers $m_{i}-m_{j}, m_{i}+m_{j}$ is not divisible by 4, and the other is divisible by $2^{i-2}$. Otherwise, $N$ is not divisible by 5, and similarly, we obtain that one of the numbers $m_{i}-m_{j}$, $m_{i}+m_{j}$ is divisible by $5^{i-1}$.

Consider a five-element subset $\tilde{A} \subset A_{1}$, denote the smallest element of $\tilde{A}$ by $u$, and the largest by $v$. Let $r=\lceil(u-1) / 2\rceil$. If $N$ is odd, set $\alpha=u-2, \beta=r$; otherwise, set $\alpha=r, \beta=u-1$. From the above, it follows that the elements of the set $\left\{m_{s}: s \in \tilde{A}\right\}$ give no more than two different remainders modulo $2^{\alpha}$ and no more than two different remainders modulo $5^{\beta}$. Therefore, in $\tilde{A}$ there will be two different elements $i10^{u-1} 2^{u / 2}
\end{aligned}
$$

from which it follows that $v / u>1.01$. Thus, if we divide the interval $[k / 2, k]$ into groups of consecutive numbers, in each of which the ratio of any two elements is less than 1.01 (the number of such groups is less than, for example, a million), then each of these groups contains no more than 4 elements of the set $A_{1}$. From this, it follows that the number $\left|A_{1}\right|=f(N)-1$ is bounded.