Berruv S.l.

On the board, 100 pairwise distinct natural numbers $a_{1}, a_{2}, \ldots, a_{100}$ were written. Then, under each number $a_{i}$, a number $b_{i}$ was written, obtained by adding to $a_{i}$ the greatest common divisor of the remaining 99 original numbers. What is the smallest number of pairwise distinct numbers that can be among $b_{1}, b_{2}, \ldots, b_{100}$?
If $a_{100}=1$ and $a_{i}=2 i$ for $i=1,2, \ldots, 99$, then $b_{1}=b_{100}=3$, so among the numbers $b_{i}$ there will be no more than 99 different ones.

We will prove that among the numbers $b_{i}$ there will always be 99 different ones. We can assume that $a_{1}a_{j} \geq a_{i}+d_{i}=b_{i}$, from which $b_{i} \neq b_{j}$. Thus, all 99 numbers $b_{i}(i \neq k)$ are different.

Second method. Let $c_{i}=a_{i+1}-a_{i}$. Let $c_{m}$ be the minimum of the numbers $c_{1}, c_{2}, \ldots, c_{99}$. We will prove that $b_{i}<b_{j}$ if $i<j$ and $i \neq m+1 \neq j$. We will consider two cases.

1) $i<j-1$. Then $i \leq 98$, and the numbers $a_{i+1}$ and $a_{i+2}$ are divisible by $d_{i}$. Therefore, $d_{i} \leq a_{i+2}-a_{i+1}<a_{j}-a_{i}$, from which $b_{i}=a_{i}+d_{i}<$ $a_{i}+\left(a_{j}-a_{i}\right)=a_{j}<b_{j}$
2) $j=i+1$. Then $i \neq m+1$ and $i \neq m$. Therefore, the numbers $a_{m}$ and $a_{m+1}$ are divisible by $d_{i}$. Hence, $d_{i} \leq a_{m+1}-a_{m}=c_{m} \leq c_{i}$, consequently, $b_{i}=a_{i}+d_{i} \leq a_{i}+c_{i}=a_{i+1}=a_{j}<b_{j}$.