Can you just say that the combination can make an addition of $-1$ and $1$, so $\forall n \epsilon \mathbb{N}$ it can make $2020$

YerMom wrote: Can you just say that the combination can make an addition of $-1$ and $1$, so $\forall n \epsilon \mathbb{N}$ it can make $2020$ Yes, and in fact, Kosovo problems are well-know to be trivial

Math-Shinai wrote: YerMom wrote: Can you just say that the combination can make an addition of $-1$ and $1$, so $\forall n \epsilon \mathbb{N}$ it can make $2020$ Yes, and in fact, Kosovo problems are well-know to be trivial https://artofproblemsolving.com/community/c6h1821954p12173226

Math-Shinai wrote: YerMom wrote: Can you just say that the combination can make an addition of $-1$ and $1$, so $\forall n \epsilon \mathbb{N}$ it can make $2020$ Yes, and in fact, Kosovo problems are well-know to be trivial Well in Kosovo the level of participants are not to high and if you will see the result you will be surprised that even with this kind of problems score to low. Even in this problem you will be surprised if I tell maybe nobody could solve this. So we try to do problems that at least somebody could solve and then those that have some talents to advanced in more knowlege and in those last years we are training participants also. I know maybe we are to far to perfection but at least the TST on last years especially last year were pretty good and not trivial at all.

If Ben decides to take subtract 2 times and add once, we have: $n-4+3=n-1$ And if Ben decides to subtract once and add once, we have: $n-2+3=n+1$ Since we have shown that it is possible to add or subtract 1 from any given number, it is certain that through an order of moves, Ben can eventually reach 2020 from any natural number $\blacksquare$