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42

Move at random. Any deterministic strategy you choose has a chance that your partner will choose the exactly opposite strategy, so you end up moving along more or less antipodal paths and never meet. So deterministic strategies have to be avoided. You might make some adjustments to your random strategy. For example, you could prefer to walk longer ...


33

As per "You have been put on the sphere without being able to communicate a plan." I'm going to assume you cannot even assume what plan your partner may come up with and there is no prior collaboration. Given the potential symmetric nature of the problem, there must be a random element to break that symmetry, should you both accidentally choose mirror ...


12

move on spirals like this: where the distance between spiral arms is $2\epsilon$. I assume you have a measure of distance on your sphere, or else you couldn't determine the winning condition.


7

Here there are some points. Here it seems a bit redundant to me to talk about strategies, because it is more of a static game with complete information than a dynamic game. First of all, I assume that for you this is a game with complete information. To do so more explicitly, we ā€“ roughly ā€“ simply have to add to your sentence "There are no visual cues on ...


6

Since you've included stars such that can be navigated from in the problem statement, this means that at least one unique "configuration" of star positions is visible from any given point on the sphere. From this, I would posit that a better-than-random solution would include finding the most "interesting" such configuration (so first you have to map them ...


6

An excerpt from "C21 Sums of higher powers" in Richard Guy's Unsolved Problems in Number Theory (third edition, pg. 207): Denote by $s_k$ the largest integer that is not the sum of distinct $k$th powers of positive integers. Sprague showed that $s_2=128$; Graham reported that $s_3=12758$ and Lin used his method to obtain $s_4=$ .... Here's the ...


5

Note that all the elements of the set are odd. Hence, even if we repeat, we have, w.l.o.g, the following cases (i) Odd $x$ + Odd $x$ + Odd $y$ = Even+Odd=Odd (ii) Odd $x$ + Odd $y$ + Odd $z$ = Odd (iii) Odd $x$ + Odd $x$ + Odd $x$ = Even+Odd=Odd I actually remember this question which is rumored to be from an IAS exam and no one could ...


5

I have listed all possible times on a $24$-hour system that appear as palindromes on a digital clock. You may need to ignore the colon ($:$) in order to regard it a palindrome. For example, $\text{2:52}$ is read backwards $\text{25:2}$, but ignoring the colon gives you the palindrome number $252$. $$\begin{array}{|c|c|c|c|} \hline \text{0:00} & ...


5

I think completely random movement is sub optimal. I think a better strategy is to pick a direction (any will do), stick to it and randomise your speed. If both parties do this their paths will cross twice each orbit (unless they are on the same orbit in which case they will meet sooner due to randomised speed) if you go full random changing direction as ...


5

$$ \begin{align} &1+2+\cdots+2015+2016+\cdots+4030\\ =\:&S+(2015+1)+(2015+2)+\cdots+(2015+2015)\\ =\:&S+(1+2+\cdots+2015)+(\underbrace{2015+2015+\cdots+2015}_{2015\text{ times}})\\ =\:&S+S+2015\cdot2015\\ =\:&2S+2015^2 \end{align} $$ If you want to find the value of $S$, you can do the following additional pairing: $$ \begin{align} ...


4

Your equations are correct, so you apparently just made a mistake in solving the system. Iā€™d multiply the second equation by $10$ to get $200=b+0.2g$ and then subtract that from the first equation to get $200=0.8g$. Then $g=250$, so $b=400-250=150$.


4

$$\def\r{\color{red}{1}}\def\b{\color{blue}{2}}\def\g{\color{green}{3}} \matrix{\r\cr \r&\r\cr \r&\r&\r\cr \r&\r&\r&\r\cr \r&\r&\r&\g&\g\cr \r&\r&\r&\g&\g&\g\cr \b&\b&\b&\g&\g&\g&\g\cr \b&\b&\b&\g&\g&\g&\g&\g\cr}$$ can be rearranged as ...


3

Are you allowed to leave your 'footsteps'? I believe it makes the problem more interesting (otherwise the solution of MvG seems to be correct). My solution is based on the fact that 2 great circles must intersect on two points (or be the same circle). So I suggest that both persons will start walking in a great circle. Eventually, each one will reach an ...


3

Suppose that your drawing DOES have a path that does what you say. Any vertex is either a start or end vertex or an interior (visited along the way) vertex. If it's an interior vertex, it has an EVEN number of edges meeting it, because for each visit, there's the edge you arrived on and the edge you left on. For a start/end vertex, it has an ODD number ...


3

This is a problem about knowledge, not about probability. When the traveller leaves, they all know that at least one dragon has green eyes. If there had been only one such dragon, that dragon would know they had green eyes and would transform at midnight. The next morning all the dragons know that no dragon transformed at midnight. They therefore also know ...


3

July 16th. Albert says that he knows that Bernard doesn't know when Cheryl's birthday is. If her birthday was in the month of May or June, then it's possible that it could have been May 19th, in which case Bernard would know from the date alone what the birthday was. Likewise, it could have been June 18th, and Bernard would have known from the date alone. ...


3

Hint: suppose the first player takes $5$ matches on his first turn. Then there are $70$ matches left, and $70$ is divisible by $7$...


3

This is four successive applications of the pigeonhole principle (PHP). We can do it various orders, but I will first look at weights within a colour. If any five balls contains two of the same weight, then there are at most four different weights of ball of any colour, by PHP. Then we look at the number of balls in total. With 49 balls, we must have more ...


3

If you add the $1$'s you get fouR If you add the $2$'s you get eighT If you add the $3$'s you get twelvE


3

Idea In a vector space of dimension $6$, any $7$ vectors are linearly dependant. Details The $63$ cards can be represented as the elements of $\mathbb F_2^6 \setminus\{\mathbf 0\}$: The six coordinates correspond to the six colors. If a color is present on a card, the corresponding entry is $1$ and otherwise $0$. Now let $S \subseteq \mathbb F_2^6 ...


3

Define $S_0=0$ and consider the partial sums $S_k:=\sum_{i=1}^k a_i$. Let $k^*$ be any value of $k$ where $S_k$ is minimum. If $k^*=n$, set $k^*=0$. Then the rotation $[a_{k^*+1}, a_{k^*+2}, \ldots,a_{k^*}]$ should satisfy your criterion. To see this graphically, create a second copy of $[a_1,\ldots,a_n]$ and append it to the original, then plot $S_k$ vs ...


2

This can be solved by the following greedy strategy (where we assume each ribbon has a unique color and hence a unique target, as all cases may be reduced to this one). The main body of the post proves it solves things in at most $n+m-1$ moves. The addendum proves that it is optimal: Fill the grid from top to bottom (i.e. starting at the topmost cell, ...


2

Let $L$ be the proposition that the left door leads to escape. You ask a question $Q$ (some proposition). Let $R$ be the truth-telling/lying status of the person you ask (so true if the person is a truth-teller, false if the person is a liar). The response from this person is the truth value of $Q \Leftrightarrow R$. So what you want to do is come up with ...


2

Waring's problem gives $g(4)=19$, i.e., every positive integer can be written as the sum of $19$ fourth powers, and this is minimal. So there is no largest number, which cannot be written as the sum of fourth powers. Edit: The question is about the sum of distinct fourth powers. Then there are some numbers which are not the sum of distinct fourth powers. ...


2

Total apples $10 + 12 + 15 + 20 + 22 + 25 = 104$ The problem is basically saying that $104 - y = 0 \mod{3}$ This is because we know that $2g = r$ and therefore the total apples must have been divisible by $3$ to get an integer value of $g$ and $r$. So when you subtract each basket, you need to see if the result is divisible by $3$. When $y = 20$, we have ...


2

Let's suppose that your opponent's strategy is to redraw on $r$ or lower. If your first draw is $x$, then the probability you win if you keep that draw is $$ P(\text{oppenent's final score}\le x)=\begin{cases} rx & \text{if }x\le r \\ rx+x-r & \text{if }x\ge r \end{cases} $$ This is because, when $x\le r$, the only way to win is if your opponent ...


2

I assume that the coin is tossed each trip only upon reaching the wall position, and also that the diagram is a top view so we are walking around, not climbing over, the wall. Firstly, the default expected distance should probably assume, in the case of tails, that you walk diagonally on the second part of the trip, so we would have $E(D) = \frac{1}{2}2 + ...


2

You have $$4^{11}=4194304\equiv 4 \pmod{100}$$ And since $2015=183\times 11+2$, and $183=16\times 11+7$, you have $$4^{2015}=4^{11\times183+2}=4^2\times(4^{11})^{183}\equiv 4^2\times 4^{183} \pmod{100}$$ $$=4^2\times4^{11\times16+7}=4^9\times(4^{11})^{16}\equiv 4^9\times4^{16}=4^{25}=4^3\times(4^{11})^2\equiv4^5 \pmod{100}$$ $$=1024\equiv 24\pmod{100}$$ ...


2

$$ 4^{2015}\cdot9^{2016} = 9\cdot36^{2015} $$ So look at $9\cdot36^n$ for $n=1,2,3,\ldots$: \begin{align} 9\cdot36^1 & = 324 \\ & = \cdots24 \\[6pt] 9\cdot36^2 & = \cdots64 \\ 9\cdot36^3 & = \cdots04 \\ 9\cdot36^4 & = \cdots44 \\ 9\cdot36^5 & = \cdots84 \\ 9\cdot36^6 & = \cdots24 \longleftarrow\text{Now we're back to where we ...


2

The cards correspond to (nonzero) vectors in the $6$-dimensional vector space $V$ over the $2$-element field ${\Bbb F}_2$. Under this correspondence, a "set" is a collection of vectors whose sum is the zero vector. Since the dimension of the vector space is $6$, any set of $7$ vectors $v_1,\ldots, v_7$ must be linearly dependent, i.e. there exist constants ...



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