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12

I have a "magic trick" I like to perform for classes on a random day throughout the semester (I cannot take credit for its creation! I do not know the original source...). "Magic" trick: Pick a four digit number with at least two distinct digits (four is completely arbitrary). Rearrange this number any way you like, swapping at least one pair of digits ...


11

Proof that $N=49$ is impossible: +-------+--- 1) +-------+ 2) +-------+ 3) +-------+ | 1 2 3 | X |(1)2 3 |*1 | 1 2 3 |(7) | 1 2 3 |(4) | 4 5 6 | Y |(4)5 6 |*4 | 4 5 6 | 1 |(4)5 6 | 7 | 7 8 9 | Z |(7)8 9 |*7 |(7)8 9 | 4 | 7 8 9 | 1 +-------+--- +-------+ +-------+ +-------+ ...


7

We see that we are trying to find the probability $$P(A\text{ fails before time } t \wedge (B\text{ fails before time } t \vee C\text{ fails before time } t).$$ By independence, this is equal to $$P(A\text{ fails before time } t)P(B\text{ fails before time } t \vee C\text{ fails before time } t),$$ We will label these events as $A_t,B_t,C_t$ for ...


6

Kaprekar's constant Start with any number of up to $4$ digits, with at least two different digits: add leading $0$'s if necessary to make $4$ digits. Write the digits in decreasing order and in increasing order, and subtract the second from the first. Repeat $7$ times. The result will be $6174$. For example: start with $1234$ $4321 - 1234 = 3087$ ...


5

You should stay with your initial choice until $N-2$ doors have been opened. Then switch to the single door you can switch to. With this strategy you are sure to win except when your initial choice happened to be the prize door. In other words, your chance of winning is $\frac{N-1}{N} = 1-\frac1N$. If you switch any earlier, your chance of having the ...


3

There's a subtle flaw in your reasoning. You're effectively conditioning on which non-common gem type comes first. Your calculation for phase $2$ is correct: The non-common gem type already seen in phase $1$ has become irrelevant, and only the relative probabilities of the other two gem types matter. But this doesn't work in phase $1$, because the length of ...


3

Hint: We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required. You can do ...


3

We first go after the complementary event, the event the system is still alive at time $t$. This event can happen in two disjoint ways: (i) $A$ is alive or (ii) $A$ is dead but $B$ and $C$ are alive. The probability of (i) is $e^{-\alpha t}$. The probability of (ii) is $(1-e^{-\alpha t})e^{-\beta t}e^{-\gamma t}$. Thus the probability the system is dead ...


2

Pick any number between 100 and 999. Write it again next to the number, and now you have a 6 digits number. Divide it by 7, divide it by 11, divide it by 13. I no step there should be any remainder (you can simulate that you're guessing or thinking very hard about them). The end result is the number chosen initially. For example for 439: 439439 (write it ...


2

The limit you took doesn't correspond to any limiting distribution. For each $x$, each occupant is assigned one of the rooms with the same probability $\frac1x$. But infinitely many occupants cannot be assigned rooms with the same probability "$\frac1\infty$". Here's a distribution for the infinite case, though: Inhabitant $n$ takes one of the rooms $1$ ...


2

I think we can do it by exhaustion. Suppose $a_2=1$. Then it must be that $a_3=2$ for the sequence must include the successor of $0+1$. At this point, the pairwise sums are $(1,2,3)$, so we would need to fill in the gap with $a_4=4$, yieding pairwise sums $(1,2,3,4,5,6)$. We would need to fill in the gap with $a_5=7$, but then our pairwise sums become ...


2

UPDATE: After adding condition 3 below (only 3 out of every 4 locations in a 2x2 square can be a block, savile row + lingeling (a SAT solver) prove there is no solution with 34 or more blocks in about 10 minutes. This does not of course provide any kind of nice proof I have tried modelling this problem in the Savile Row system, which can generate input for ...


2

Using @Emisor's construction, we can reduce the upper bound to 39. First a bit of set-up: I'll use Cartesian coordinates to refer to positions on the board, with (1,1) at the top left and (9,9) bottom right. I will identify solution blocks by the coordinates of their top left cell. I'll say the block is "anchored" at these coordinates. The obligatory ...


2

Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$


2

We can analyze this question using boolean algebra. Let $p$ represent whether the native is type A (true) or type B (false), and let $q$ represent whether Thomas is sane (true) or insane (false). The statement "Can the native ask the question 'Is Thomas insane?'" is logically equivalent to $p\oplus q$, where $\oplus$ is XOR or exclusive or. The statement ...


2

Oops: my previous analysis was wrong. A type A native asks a question for which the correct answer, not necessarily the answer that will be returned, is Yes. If a proposition $Q$ is true, a sane Thomas believes it, and an insane Thomas does not. If $Q$ is false, a sane Thomas does not believe it, and an insane Thomas does. So: a type A native can ask ...


2

Your reasoning for Case 4 might be flawed: If Thomas is insane, then his answer to the question "Are you insane" must be "No". For this reason you could consider this the correct answer to the question, in which case there is no contradiction.


1

I believe you are Correct. There are 3 lines separating rows horizontally, 4 pairs of numbers across each line, and 2 ways to combine said pairs (top-down or bottom-up), giving $2*3*4=24$ ways of making pairs. Repeating for the columns and getting the exact same numbers, we now have $24+24 = 48$ total options for merging two numbers.


1

First find out how many of the 31 coins (1 + 2 + 4 + 8 + 16) weigh 11 gm. You will find there are 13 of them. Then think of how you can write 13 as a sum of powers of 2 - this is "writing 13 in base 2". Answer, 8 + 4 + 1. So the coins that came from bags 1, 2 and 4 weigh 11 gm each, and the others 10 gm each.


1

Related to joriki's answer, if Inhabitant $n$ goes to a room between $1$ and $n\log n$, each room is also almost surely occupied, but if they go to a room between $1$ and $n^2$, the probability is less than $1$.


1

Player $A$ strikes out $21$ (so that the sum of all the remaining numbers is divisible by $5$), and then uses the following pairing: $(1,19),(2,18),(3,17),(4,16)$ $(6,14),(7,13),(8,12),(9,11)$ $(5,20),(15,10)$ Whatever $B$ strikes out, $A$ strikes out the other number of the pair, leaving a sum divisible by $5$.


1

If you were lucky enough to choose $y\geq S$, the distance will be $2\sqrt{1+y^2}$. Great. But if you weren't, the distance will be, as you said: $$d=\sqrt{1+y^2}+S-y+\sqrt{1+S^2}$$. Since this depends on $S$, the distance you will have to walk is itself another random variable. You didn't specify this in your statement of the problem, but I'll assume you ...


1

I would comment, but I don't have enough reputation. The only problem I see is you mixed up your probabilities for $P(y<S)$ and $P(y>S)$. They're actually reversed; in fact, $P(y<S)=1-y$ and $P(y>S)=y$. If you switch them, I believe the formula works fine; e.g. it gives $2\sqrt 2$ for $S=0,y=1$.


1

Too long for a comment: Certainly no choice can be required here. The comments sketch a solution, but even without knowing the solution, this follows from a general fact: Shoenfield absoluteness. Shoenfield absoluteness (among other things) isolates a class of statements which, if provable from ZFC, are also provable from ZF. Certainly not every statement ...


1

The shortest time solutions that I am aware of are based on this scheme, first suggested by Paul Hammond here. Time is broken into Rounds, and each round is further broken into two Stages. In Paul Hammond's original scheme, the first stage in each round is 2600 days, while the second is 2700 days. Each prisoner is considered to start with a virtual ...



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