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Let the combining operation be $\otimes$, i.e. $a \otimes b = ab + a + b$. Observe the following for $\otimes$: $$a \otimes b + 1\\ = ab + a + b + 1\\ = (a + 1)(b + 1)$$ So the ordinary multiplication operation is equivalent to $\otimes$ just offset by $+1$. This suggests an equivalent formulation of the problem where instead of working on the original ...

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First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$. Now choose $2015$ and $2013$. Average = $2014$. Choose $2014$ and $2012$. Average = $2013$. Note that we can keep on continuing this approach and end up with a situation like $1$, $2$ and $4$ in the end. From here, choose $2$ and $4$. Average = $3$. ...

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Dr Xorile's answer of is optimal. Suppose that the numbers 1, 2, 4, 8, ... 512, and 1024 are all on the board. In order to eliminate the number $2^i$, there must be some step where you decrease $2^i$, but none of the numbers $2^0,2^1,\dots, 2^{i-1}$. If this were not true, then the amount you decreased $2^i$ would be at most the amount you decreased the ...

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It can be done in: "But how?" you ask. Well, I'm glad you asked: Convert the numbers to binary, and subtract the bits. Or, to put it another way, subtract 1024 from all the numbers that can afford it. Then subtract 512 from all the numbers that can afford it. Then 256. Then 128. Etc. Down to 1. This is the best answer. See @Mike Earnest’s proof ...

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The largest it can be is actually also the smallest it can be. In fact, if the numbers 1 through $n$ are written and the same process followed, the end result will be $(n+1)! - 1$ no matter what order you combine numbers. Let's take a smaller set, just $\{a, b, c\}$, to see why. If you group $a$ and $b$ first, you'll end up with $$(ab+a+b)c+(ab+a+b)+c=a+b+... 20 Parity If you remove any two even numbers, their difference will also be an even number. The total number of odd numbers will remain the same. If you remove any two odd numbers, their difference will be an even number. The total number of odd numbers will decrease by two. If you remove an even and an odd number, their difference will be an odd number. ... 19 Notice that$$\frac{1}{\frac{xy}{x+y}}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}$$So the sum of the reciprocals of the numbers on the board is always preserved. The sum of the reciprocals of the numbers in each row is 1, so the sum of all the reciprocals is 10. Therefore, the last number left must be \frac{1}{10}. 18 Building on @2012rcampion's answer, here's a simple constructive proof that any even number is reachable. Take the (even) number that you want to reach (say n) and put it aside. Then take the difference between all the pairs (500,499), (498,497),\ldots,(n+2,n+1), and (n-1,n-2),(n-3,n-4),\ldots,(3,2). That will give you an odd number of 1's. You'... 17 Let A(x)=2-\frac{1}{x} and B(x)=1+\frac{1}{x-1}. Notice: In total, the sequence of button presses is (One continued fraction representation of \frac{1003}{100}-1 is [9;33,2,1]. This is not a coincidence, which can be explained if you consider how the operations affect x-1.) 13 Answer The (maybe) interesting details The ugly details 12 Reasoning: After this we... ...going all the way back to... 12 Obviously we're never going to get an irrational number. So any number we get can be expressed as \frac{p}{q}, where p and q are relatively prime integers (q can be 1, if the number is an integer). When we have \frac{p}{q}, we can write \frac{2p+q}{q} and \frac{p}{p+2q}. If p and q are both odd, 2p+q and p+2q are also odd, so any ... 11 By the rational root theorem, any integer root of the polynomial$$a_nx^n + \dots + a_1x + a_0$$is a divisor of its constant term a_0, if a_0 \neq 0. We will never let a_0 = 0, since that means we can divide the whole polynomial by x (repeatedly, if necessary; yielding a root 0 each time, which is initially already on the board) and get a ... 9 A simple way to get to 1 First we will transform the squares 1 to 81 as follows: 81-64=17 49-36=13 16-9=7 4-1=3 17-7=10 13-3=10 25-10=15 15-10=5 We are left with the 5 and additional 92 squares from 10^2 to 101^2. Each consecutive 4 squares can be transformed to a 4 like in the following example using 100, 121, 144 ... 9 So far I have a lower bound of and an upper bound of For the upper bound, consider the following simple strategy: Ignore 0's If there's two n's, change them to n-1 and n+1. Otherwise, create two 1's. Let's keep ignoring 0's and pretend that we can add a single 1 for half a step. We argue that at any point, each number appears once, except at most ... 9 The smallest possible value of n is Claim: We can get every non-negative integer n\leq 2016 on the board. Proof: By induction. We start with n=0 on the board. We can get 1 using Lord of the Dark's method: 2016!x+2016!=0 has -1 as a root, -x^2+2016!=0 has \pm\sqrt{2016!} as roots, and \sqrt{2016!}x-\sqrt{2016!} has 1 as a root. Now ... 7 The answer is Proof: 6 The shortest amount of time it takes to write the number 100 is I will need to following fact. Claim: Let k be a non-negative integer. If there has ever been an integer n>k written on the board, then for any interval of k+1 consecutive positive integers less than or equal to n, there are at least k integers from that interval currently ... 6 The answer is : Explanation : Generalization : 6 Obviously we have x=2016. Let x=2, then we can write \dfrac24. With x=4 we can write \dfrac46, etc... So any even number \lt2016. 5 I wanted to post my solution as well, because it looks a little diffferent than f'''s. If we press [A] repeatedly we see \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \ldots, and indeed [A] takes 1+\frac{1}{n} to 1+\frac{1}{n+1}. This suggests looking at the quantity y:=\frac{1}{x-1}, where x is the displayed number. The conversion the other ... 5 I tried tackling the problem numerically. I found that up to 13 layers of depth(the furthest I could go), there was no solution. First I inverted  f(x) = 2x+1 -> f^{-1}(x) = \frac{x-1}{2}  g(x) = \frac{x}{x+2} -> g^{-1}(x) = \frac{2x}{1-x} Next, I want to consider all possible inversions of 2016 at a depth n. The question we are trying to ... 4 If you keep taking the largest 2 numbers, and substract them from eachother, you end up with There's no very clean way to illustrate this (that i came up with) besides giving the calculations used Where the bolded area indicates the part that's different from Joe Z's method. 3 I believe the OP's intended answer is: Exactly the divisors of 120. This is true of the non-zero numbers on the board at any stage. If P(x)=0, then we can write it as xg(x)+d=0 and d is a divisor of 120, this implies that so is x. StillLearning's answer shows that all of them can be obtained. Once 2,3,5 are obtained, other divisors of 120 can be ... 3 I started like this$$\begin{align} 120x + 120 = 0 &\rightarrow& x &= &-1\\ - x^{120} - x^{119} - ... - x^2 - x^1 +120 = 0 &\rightarrow&x &= &1\\ x + 120 = 0 &\rightarrow& x& = &-120\\ x^7 - x^3 + 120 = 0 &\rightarrow& x &= &-2\\ x - 2 = 0 &\rightarrow& x &= &2\\ \end{align} $$... 3 Every integer can be written on the board. First polynomial:$$P(x) = 120x - 120$$The root of this polynomial is 1. Any integer:$$P(x) = x - x In this solution, it is important to note that constants are not coefficients. Also note that $-$ is an operator, not part of the coefficient. Had it been intended as the coefficient, I would have written $x +... 3 If you erase$n^2$and$(n+1)^2$starting from$n = 2$and going up by$2$, you end up with the numbers$1, 5, 9, \ldots, 201$, which is a total of 51 numbers. If you then erase pairs of consecutive numbers starting from$5$and$9$, you end up with a single$1$and 25$4$'s. If you erase consecutive pairs of 4's until they become$0\$, you're left with one ...

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Partial Answer : (?) Explanation : Example :

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It can be assumed that evaluation is left to right (or rather: clockwise), not by precedence rules. Then this reads and simplifies to or

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