Eugene wrote a 100 numbers on a board. He added 1 to each number and the product didn't change. He did the same thing k times, each time the product didn't change. What is the maximum k?
I guess the answer is 99 and that happens if the numbers are 99, 98, 97, 96,..., 0. But how do you rigorously prove it? Stuck on it for about an hour. Tried induction with 2, 3, 4 numbers
Let the $100$ numbers be $a_1, \cdots a_{100}$. Consider the polynomial $(a_1+x)(a_2+x)\cdots(a_{100}+x)-a_1a_2\cdots a_{100}$. This has at most $100$ roots. Therefore, there can be at most $100$ values of $x$ for which the product is the same as it is initially, including at time $0$. It follows that the value of $k=99$, with the initial numbers of $-99,-98,\cdots,0$ is indeed optimal.
