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43

Exactly two underscores must be occupied by operators. Furthermore, no two operators can be placed adjacently, as this would be syntactically meaningless. Hence, there are only three possible arrangements of digit and operator positions: $2$ # @ # @ $5$ $2$ @ # @ # $5$ $2$ @ # # @ $5$ For each case, we choose the digits and operators that give the ...


37

This is not possible, the largest number we can make using two operations and two digits is given by $$ 2 \cdot 9\cdot 95=1710<2015$$ so this problem has no solution stated as such. Of course if you relax the condition 2 operations-2 digits then, as proposed by 5xum (who proposed 2010+5), it is much easier.


36

No where in the puzzle forces us to interpret the digits as decimal numbers. In base $8$, we have: $$21 \times +75 = 2015$$


22

The puzzle does not restrict the operations to binary, so using + as a unary operator opens up even more possibilities and allows to reach higher numbers. It seemed promising at first, but sadly none of the possibilities is 2015. Here's a table: $$ 2\mathrm{X} \times +\mathrm{Y}5$$


5

Let's assume $K_4$ instead of an actual tetrahedron, since I'm envisioning some serious avalanches here. :-) There are six roads in all. If at most two roads are open, then at least one village cannot be reached, since two roads can only connect three villages. If at most two roads are closed, then all villages can be reached, since no two roads can ...


5

Brute force solution I originally wrote the solution for 5 slots, with only 4 slots this approach proves that only two solution candidates appear: 2010+5 and 2020-5. Obviously these don't match the criterium so it is proved that no solution exists. For illustration I will hereby show how I found the solution for when there are 5 slots. Approach It can ...


4

As shown by the other answers, this problem is impossible with two operators and two regular digits. So we'll have to take liberties: Taking the liberty of an extra possible operator gets us closer, as shown by Surb's comment to his own answer. If we take another operator than what he suggests, and include the factorial operator, then we get very close: ...


4

Let the question be $x + y + z = abcdef$. Throughout, multiplication will strictly be represented by $\times$, and concatenation of variables corresponds to concatenation of digits. As you note, $x \times y = ab$. $cd = x+y+z+a+b$. $fe = ab + cd - z$.


4

How about this?$$\dfrac{6}{1-\dfrac 34}$$


3

Like it usually happens with these conjectures, if the solution is not k=7 or k=11, it may be really hard to find. Of course, there exists a solution, for there is a maximum (huge) number of solutions that a blank sudoku allows. A property that may help to construct solutions is that for small values, if we have a sudoku that allows a solutions and another ...


3

This is a weird outfit for a Raymond Smullyan puzzle to be wearing. The first native, whether a Bear or a Packer, will invariably deny being a Bear. Thus, the second native is telling the truth and is a Packer. The third native's claim that the first native is a Bear could be the truth, in which case the first native is a Bear and the third is a Packer; ...


3

Tetration notation: _ _ _ |_| _| |_| _| | _| |_| _| _ _ |_| _| |_| _ | _| |_||_| | or (if that's not a valid $9$), _ _ |_| _| |_| _ | _| |_| _| _| $^{438}3= 3^{3^{\cdot^{\cdot^{3^3}}}}$ (an exponential tower with $438\ \ 3$s).


3

If there is no guarantee where the darts land on the dartboard, then there is no finite answer to the question. (See SE318's answer) If you know that the first $N$ darts all fall at least $r$ distance apart from one another, then the question is equivalent to packing circles of radius $\frac{r}{2}$ inside a larger circle of radius ...


3

Assume that you have an algorithm to find the first three places in 6 races. Assume that after the fifth race $n$ horses have raced. If $n=25$, then the last race can detemine only the winner, but not the second nor the third place: it has to race one horse from each previous race, so if the first three places were all in the same heat or 1,2 in the same ...


3

The optimality of the strategy is the consequence of the Bregman-Minc inequality. It says that the permanent of a matrix $A = (a_{i,j})_{i,j=1}^n\in\{0,1\}^{n\times n}$ with $r_i = \sum_{j=1}^n a_{ij}$, $i=1,\dots,n$, satisfies $$ \operatorname{perm} A \le \prod_{i=1}^n (r_i!)^{1/r_i}; $$ the equality holds iff $A$ is, up to permutation of rows and columns, ...


3

This isn't really an answer, just the results of some initial investigations. Let $f(n)$ be the minimum number of weighings required, and $g(n)$ be the minimum number of weighings under the further restriction that the weighings are fixed (not dependent on the weights of previous weighings). Clearly $f(n)\leqslant g(n)$. Here are some values and bounds for ...


3

The Blue Eyes Puzzle is a very good candidate in my opinion.


2

This might be a cliché by now but I've always liked the locker problem.


2

I think that many combinatorics problems will be a good candidate, because counting is very intuitive to everyone (and there are very few theorems to be used). A personal favorite is the following: Let $a_n$ be a nondecreasing sequence of integers, and $b_k$ be the number of terms of {$a$} that are greater than $k$. Prove that the sum of the elements of the ...


2

This is a partial answer, identifying some $n$ that cannot be synthesized with a finite subset of values from $E$. The LCM of $E_{\text{b}}$ is $N=3541509972=2^2\cdot 3^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 41\cdot 47$. Thus $\frac{N}{e}$ is an integer for $e\in E_{\text{b}}$. For general $e\in E$, $\frac{N}{e}$ therefore has fractional part with finite ...


2

Yes, it is the $k=2$ case of the Josephus problem, but you are off by one from the classic formulation because you kill the first instead of the second. Take the classic solution, then subtract one from the last one standing to account for that. If the classic last standing is $1$, it will be $N$ as the subtraction should be done $\bmod N$


2

I think this is a fantastic question, because you actually arrive at a pretty good strategy for how to begin a game. Sorry it's such a long answer, I try to be thorough so you can see this is the correct answer - there is a summary at the end. I'm going to rephrase your question in the way I interpret what you mean. At the start of a game of minesweeper, if ...


2

You cannot guarantee this since for example you can throw infinitely many darts right near the leftmost part of the circle and at any step you could have hit on the rightmost part giving you a distance greater than r.


2

All integers are of the form $10n\pm k$, where $0\le k\le5$. Squaring, we have $100n^2\pm20kn+k^2$. For $k^2<10\iff0\le k\le3$, the tens digit is always even. For $k=4$ we have numbers ending in $4$ and $10-4=6$ yielding squares with an odd tens digit, since $4^2={\color{red}1}6$. For $k=5$ we have $k^2={\color{red}2}5$, so the tens digit is again ...


2

The sequence $x_j = b^{2j+1} - b^{-(2j+1)} - 2 (c^{2j+1}-c^{-(2j+1)})$ satisfies the recursion $$ x_{n+4} - 9 x_{n+3}+20 x_{n+2}-9 x_{n+1}+x_n = 0, x_0 = 0, x_1 = 6, x_2 = 60, x_3 = 420 $$ I looked at this modulo each of the primes $17, 31, 41, 43, 89, 97, 167, 331$ that divide your $a = 443372888629441$, and in each case $x_{(a-1)/2}$ is congruent to $0$. ...


2

This is just a partial answer, which can perhaps be turned into a community wiki, where examples of partial fillings that allow $n$ complete valid fillings. This admits $k=2$ solutions: \begin{array}{ccccccccc} . & . & . & . & 5 & 1 & 8 & 7 & 6 \\ . & 7 & . & 6 & . & 9 & . & . & 3 \\ 6 & ...


1

The question "Can we find $1000$ numbers in the range $[0,2^{n}-1]$?" can be reworded as "Can we find $1000$ subsets of $\{1,\dots, n\}$ such that There do not exist distinct $A,B,C,D$ with $A \cup B=C \cup D$ There do not exist distinct $A,B,C$ with $A \cup B=A \cup C$ In extremal set theory, such a collection of subsets is called a strongly union free ...


1

For the second part: Since $\sigma$ is increasing, it suffices to bound $\sigma(n)$ for $n = 10^m$, $m\in\mathbb{N}$. It is not too hard to see that $|S(10^{m})| = 9^{m}-1$, so if we let $T_m = S(10^{m+1})\backslash S(10^m)$, then $|T_m| = 9^{m+1} - 9^{m}$. We have \begin{align*} \sigma(10^{m}) = \sum\limits_{k\in S(10^m)}{\frac{1}{k}} &= ...


1

Hint for the second part. Consider the number of elements of $S(10^n)$ larger than $10^{n-1}$, and observe that they all contribute less than $1/10^{n-1}$. Obtain the appropriate geometric series with ratio $9/10$.



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