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34

Subtracting the month and day lets you take into account whether you've had your birthday this year. If you have had your birthday, then you can just skip the month and day, they don't affect the two years at all (do the subtraction by hand to see this more clearly). If you haven't had your birthday yet, then the process of borrowing will subtract a year for ...


16

$$\large\begin{align*} \hphantom{-}(\mathsf{year}_1)(\mathsf{month}_1)(\mathsf{day}_1)\\ -(\mathsf{year}_2)(\mathsf{month}_2)(\mathsf{day}_2) \end{align*}$$ is equal to $$\large\begin{cases} \hphantom{{}-1}(\mathsf{year}_1-\mathsf{year}_2)\Box\Box\Box\Box&\;\textsf{if }(\mathsf{month}_1)(\mathsf{day}_1)\geq (\mathsf{month}_2)(\mathsf{day}_2)\\ ...


12

I'm leaving the intermediate steps here so you can see how the solution developed; to summarize: The result is a hypothesis for a complete solution based on strong numerical evidence but so far without an idea how to prove its optimality. For all $n$, the solution consists of starting at some point up to $3$ slots away from (say, to the right of) the centre ...


8

For most positions of the holes there seems to be room for 10 rooks. If the holes are in the same row (not on the edge, not touching each other, and with at least two rows above and below it), then place 7 rooks as R R R o R o R R R There are then 3 rows and 3 columns left without anything in them yet; put the last 3 rooks there. If the ...


7

Update: I've now found two solutions that require only $25$ weighings: $$ 01011011100010111101000001100111110011010100011010\;,\\ 00101110001101001010111110001000110010011111011101\;. $$ These are the only solutions known (in the context of this post) with $k=n\,/\,2$, for any $n$. I found them using Tad's recipe of repeatedly maximizing the determinant of ...


7

The number of cubes with at least one black face is $9^3 - 5^3 = 604$. Following picture is a visual illustration of the configuration of cubes. Cubes with at least one black face has been make partially transparent. As one can see, there are $(2 + 1 + 2)^3 = 5^3$ cubes whose faces are all pink.


6

The following pattern works with $31$ weighings: $${0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, \ 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, \ 0, 1, 1, 1}.$$ This beats the $35$-weighing solution found earlier. $${1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, \ 1, 1, 1, 0, 0, 0, 1, 0, ...


5

Let’s look at the non-negative integers having base ten representations requiring at most $d$ digits. (Including $0$ makes no different in the limit and makes the calculation simpler.) There are $10^d$ of them. Of those, $9^d$ have representations without a $7$. Thus, the fraction of these numbers with a $7$ in their representations is ...


5

Here are some lower bounds; I computed these by simulated annealing. The notation is best understood by thinking about the problem as follows: at each step, you pick a nonzero number in the line, add it to its nearest nonzero neighbors, and zero it out. So, for $n=3$, a game might run $$ (3,2,1,2,3) \to (3,3,0,3,3)\to (6,0,0,6,3)\to ...


4

Define the function $f:S^1\to \mathbb{R}$, where $S^1$ is the unit circle (in other words all possible directions on the plane). For a fixed $x\in S^1$ define the rectangle $R(x)$ as follows: there are to lines $l_1$ and $l_2$ that have the same direction as $x$ and are touching our curve, also there are two lines $l_3$ and $l_4$ that have perpendicular ...


4

This is an elementary approach, but for larger numbers, it is more computable than the total approach given by Tad. So probably this is far from being optimal: If you first weigh the coins $0$, $12$ and $25$ it takes 38 weighings according to the rule, so you can tell all coins apart. Proof: Set $w_i$ the weigh of the $i$'th coin, and $M_j$ the result of ...


4

The painting of the $27$ middle-sized cubes produces $27\cdot8$ small cubes with a black corner, $27\cdot12$ small cubes with a black edge and $27\cdot6$ small cubes with a black face (and leaves $27$ small cubes in the centres all pink). The painting of the big cube destroys $8$ corners. Along each of the $12$ edges of the big cube, there are $3$ small ...


3

So I think this will work: $$ Y = \{a, b, c, d, e, f, abce, bcdf, acde, bdef, acef, abdf, abcdef\} $$ That is, $Y$ consists of all one-letter sets, the four-letter set $abce$ and its "rotations", and the six-letter set $abcdef$. That no $y_1, y_2 \in Y$ admit $y_1 \cdot y_2 \in Y$ can be seen as follows: A one-letter set combined with a one-letter set ...


3

We can get an algorithm to determine the maximum number of rooks by considering attacks along ranks (parallel to one axis of the board) separately from attacks along files (parallel to the other axis). So first consider only attacks along ranks. A rank with a single hole in it can still hold only one rook if the hole is at either end of the rank, but can ...


3

Here is the drawing I meant to make for ten rooks.(holes are yellow and rooks are black)


3

You just have to check two cases, if $mmdd$ of the day you where born is more than current $mmdd$ then the first $4$ digits will be $yyyy-1$. If $mmdd$ of the day you where born is less or eqaul to current $mmdd$ then you get $yyyy$ as answer. Which is correct in both cases.


3

One elementary fact is key to making these problems tractable: If you pass through a given square, you must cross its northeast/southwest diagonal in exactly one place. So, consider the upper-left problem. The number of allowed paths is equal to the total number of paths (on the original grid), minus the number of paths that pass through either of ...


3

Using this cryptogram solver, I get the solution "Freddie Starr ate my hamster".


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).


2

It has been pointed out in the comments that the problem is equivalent to the following: find a subset $Y$ of the vector space $V=\mathbb{F}_2^6$ such that $|Y|=13$, and $Y\oplus Y=V\setminus Y$. Necessarily, the elements of $Y$ must be linearly independent (otherwise you could only get $32$ linear combinations, much less $50$ sums.) The conditions on $Y$ ...


2

As $s(a_5,n)$ is an integer (though it depends also on $a_1, a_2,a_3,a_4$) the claim that $s(a_5,n)\to 0$ implies that $a(a_5,n)=0$ for $a_5$ large enough (depending on $n$). This is certainly false: While you didnt't specify what would count as a valid expression, I assume that $a_1-a_2-a_3-a_4+a_5$ is among them; and for $a_5\gg n$ we can always find ...


2

HINT: ‘Untwist’ the board by converting this: $$\begin{array}{|c|c|c|} \hline \color{blue}0&3&\color{brown}6\\ \hline 5&&1\\ \hline \color{blue}2&7&\color{brown}4\\ \hline \end{array}$$ to this: $$\begin{array}{|c|c|c|} \hline \color{blue}0&1&\color{blue}2\\ \hline 7&&3\\ \hline ...


2

If I wanted to explain this "paradox" intuitively, I would say that in your first explanation, the frog knows what you are doing with the flashlight, and this allows him to avoid it. In the second explanation, the frog's moves are random, and so he is eventually (with probability one) bound to run into your light. Even though the possibility of him ...


2

There is no general formula for an $m\times n$ grid, but there is a recursive approach that allows to compute the number $C(m,n)$ of different ways to cover the grid with simple cycles for any $m,n$. First note that once you have covered the grid, it suffices to know the horizontal lines, in order to determine all the cycles. So you can assign to each ...


1

Solution follows, stop reading if you want to figure it out yourself. Order nodes so that the leaf comes first and its neighbour comes next. Claim. $1, 1, -1, -1$ cannot be balanced. Proof. Suppose that in step $0$ we have the above and in step $n$, where $n$ is minimal, we have $0, 0, 0, 0$. In step $n-2$ we must have $a, -a, a, -a$ (or the last two ...


1

Define a $6\times6$ matrix as follows $$ \mathscr{M}^{(p)}_{ij}=\begin{cases} 0&\text{if i=j}\\ 1&\text{if i-th musician has heard j-th musician play after p-th performance}\\ 2&\text{if i-th musician is yet to hear j-th musician play after p-th performance} \end{cases} $$ The goal is to make every off-diagonal entry $1$ with least $p$. ...


1

Definitions Let $R$ be the radius of the pond. Let the velocities be $v$ for the duck, and $4v$ for the fox (see diagram). Phase 1 - The Headstart As long as the duck stays with a circle of radius $\frac{R}{4}$, he can ensure that he keeps the fox as far away as possible (on a diametral line to himself) by turning in a spiral, where his maximum outward ...


1

It seems to me that the number of teams you can form would be $\min\left\{N,E,\left\lfloor\frac{N+E}{3}\right\rfloor\right\}$. Certainly you can't form more teams than this. But on the other hand you can always form this many teams by the following procedure: (1) Choose two members of type new or experienced according to which group has more members ...


1

Okay, if you lose, you end up 800 down. If you win, you end up 1600 up. So you have a lot more to gain, than to lose. Of course you don't know which one is which. But you know each chance is equal, 50/50. If the choice was between 800 and 2,400, then there'd be 800 to win or lose either way. So the choice wouldn't matter, it would be "evens". In that ...


1

Define the variables as follows: $v_d, v_s$: speed of dog, soldiers, resp. $t_f, t_b$: time for dog to reach frontline, to go from first line to back, resp. Then the dog moves 50m relative to the soldiers both ways, at two different relative velocities, so $\boxed{\dfrac{50}{v_d-v_s} + \dfrac{50}{v_d+v_s} = \dfrac{50}{v_s}}$ This simplifies to ...



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