proposed by SIR ALEX

Can someone check this because this seems horribly incorrect, but where? We note that $\mathbb Q^+$ is closed under the operations $+,\times,\div$ - i.e. if we got $-2020$, we did at least one subtraction. Now, consider the instance of the last subtraction (in the order) and reverse its order. This means that the result $\alpha$ becomes $-\alpha$. We demonstrate that all we can make sure all of its parent (we think of this as a tree, with root 2020 and making sure each node has exactly 2 children (other than the leaves)) negated, through the very obvious INDUCTION. Base Case. The direct parent. In this case, the direct child would be the result of $\alpha$ and $\beta$, and there are four possibilities (as $+$ and $\times$ are commutative: It was initially $\alpha+\beta$. In this case, we do the subtraction $(-\alpha)-\beta$ to negate the original value. It was initially $\alpha\times\beta$. In this case, we do absolutely nothing and it becomes $(-\alpha)\times\beta$, which is the negative. It was initially $\alpha\div\beta$. In this case, we do absolutely nothing and it becomes $(-\alpha)\div\beta$, which is the negative. It was initially $\beta\div\alpha$. In this case, we do absolutely nothing and it becomes $\beta\div(-\alpha)$, which is the negative. Note that no other numbers on the board have changed. Induction Hypothesis. Suppose that some parent $\gamma$ of $\alpha$ could be negated, while keeping all other numbers constant. Then, we can reverse the parent of $\gamma$ without having any other number on the board changed. Induction Step. Apply the base case to $\gamma$. Thus, we see that as 2020 is the parent of all nodes, it can be flipped.

naman12 wrote: Can someone check this because this seems horribly incorrect, but where? I think you are correct. My idea was the same, just I expressed it in a different way. We say that a number $c$ follows from number $a$ if we erase the number $a$ and another number $b$ to write $c$. If $d$ follows from $c$ and $c$ follows from $a$, then we say that $d$ follows from $a$, too, and so going on. Let $x$ be the last number which Zuming wrote as a difference of two other numbers. We can negate $x$, and by repeating the algorithm you gave, we can negate every number that follows from $x$. $-2020$ follows from $x$, implying the problem's result.

Maybe I am wrong, but how are you sure that there is at least one subtraction? Edit: Sorry, I thought we have 2020, and have to reverse it to - 2020

You're given all positive integers, and you get a negative integer.

khina wrote: proposed by SIR ALEX oh lol I thought Zuming wrote this problem

If at any point zuming did $a-b$ such that $a < b$ ,then we just reverse it to $b-a$.If at any point zuming did $a+b$ where one of them is negative (say $a$) then we do,$b-a$ or $a-b$ -the one that is positive. (this time we have $|a|$ instead of $a$) Then the absolute value of the numbers zuming had in his steps of ending up at $-2020$ will not change but this time we force all the numbers to be positive which implies we end up at $2020$ this time.

instead of relying on the idea that there has to have been at least one subtraction, regular induction also works by considering the last two numbers $a$ and $b$ (pretty much what jeff10's solution). if the final operation was multiplication/division then we can flip the sign of $a$ or $b$ (depending on which was negative) by the induction hypothesis, if it was addition then we can do the same thing, and if it was subtraction we can reverse the order of his subtraction and finish.

Let \(a_1,a_2,\hdots,a_{2020}\) be the integers. Then, we have \[f(a_1,a_2,\hdots,a_{2020})=-2020\]where \(f\) is a function that randomly adds, subtracts, multiplies and divides any two numbers from those \(2020\) integers and keeps playing around until it reaches \(-2020\). Now, our goal is to show that there exists another function \(g\) with the same rules as \(f\) so that \(g\equiv-f\), then we will be done. We show that such a function exists by replacing some operations: Whenever \(f\) does \(a-b\), replace it with \(|a-b|\) and whenever \(f\) does \(a+b\), replace it with \(|a+b|\). When we just make these changes, the multiplication and division operations that \(f\) makes will automatically change, so now we get the positive value of \(f\), or in other words, there is a \(g\) so that \(g\equiv-f\) and we are done.Let \(a_1,a_2,\hdots,a_{2020}\) be the integers. Then, we have \[f(a_1,a_2,\hdots,a_{2020})=-2020\]where \(f\) is a function that randomly adds, subtracts, multiplies and divides any two numbers from those \(2020\) integers and keeps playing around until it reaches \(-2020\). Now, our goal is to show that there exists another function \(g\) with the same rules as \(f\) so that \(g\equiv-f\), then we will be done. We show that such a function exists by replacing some operations: Whenever \(f\) does \(a-b\), replace it with \(|a-b|\) and whenever \(f\) does \(a+b\), replace it with \(|a+b|\). When we just make these changes, the multiplication and division operations that \(f\) makes will automatically change, so now we get the positive value of \(f\), or in other words, there is a \(g\) so that \(g\equiv-f\) and we are done.

We being by the following claim. CLAIM: If for some set of numbers, $-X$ is possible, then $X$ is also possible for $X>0$. Proof: Let the numbers be $\{a_1,a_2,\cdots,a_k\}$. We will prove this by Induction. 1)For $k=2$, assume $a_1\leq a_2$, then the only negative number possible is $a_1-a_2$. We can see that $a_2-a_1$ is also possible. 2) For $k=k-1$ assume that this is possible. Then let $-\mathbb{E}$ and $\mathbb{E}$ be the numbers that is derived from $\{a_1,a_2,\cdots,a_{k-1}\}$ for $\mathbb{E}>0$ 3) For $k=k$, we can see that the possible negative numbers are \begin{align*} \left\{-\mathbb{E}-a_k,-\mathbb{E}\cdot a_k,\frac{-\mathbb{E}}{a_k}, -\mathbb{E}+a_k(\text{if $a_k<\mathbb{E}$)}\right\} \end{align*}We can see that, \begin{align*} \left\{\mathbb{E}+a_k,\mathbb{E}\cdot a_k,\frac{\mathbb{E}}{a_k}, \mathbb{E}-a_k\right\} \end{align*}is also possible, since $\mathbb{E}$ can also be derived from $\{a_1,a_2,\cdots,a_k\}$. $\blacksquare$ Thus from above claim, if $-2020$ is possible then $2020$ is also possible from the 2020 positive

This problem is pretty straightforward. The main idea is the following: Claim. [Main Idea] We can guarantee that all the numbers are always positive, while keeping all the absolute values of the numbers the same. Let $a_1, b_1$ be our altered numbers and $a, b$ be the original numbers that were on the board at some point. In general, if we are guaranteed that $|a_1| = a$ and $|b_1| = b$ with $a_1, b_1$ positive, we can guarantee that the number $c_1$ produced as the result of the operation between $a_1$ and $b_1$ has the same absolute value as our original $c$ produced by the operation between $a$ and $b$, as follows: [item] [*]If $a_1 = a$ and $b_1 = b$, addition is fine. For subtraction, take $c_1 = |a-b|$, which obviously works. Multiplication and division are always fine. [*]If $a_1 = -a$ and $b_1 =b$ (or vice versa), then for addition we can take $|b_1 - a_1|$, and for subtraction we can take $a_1+b_1$. [*]If $a_1 = -a$ and $b_1 = -b$, then for addition we take $|a_1 + b_1|$ and for subtraction we take $|a_1 - b_1|$, as usual. This means that the final number in our altered system will also have identical absolute value as the original final number but be positive, from where it follows that the new final number is $2020$.