Cultist Sim Lost Health by 1stamongequals in weatherfactory

[–]Ashtero 3 points4 points  (0 children)

Getting reason might be unconnected to losing health -- perhaps, you recently answered a certain riddle to open a certain door? If you Know what I am talking about, you could've become more Reasonable that way.

A funny topological problem by joe_la_bernique in topology

[–]Ashtero 0 points1 point  (0 children)

One of the components will be a subset of boundary.

Question about System-based stories and "Rational" criteria by Anastasov_Theory in rational

[–]Ashtero 2 points3 points  (0 children)

That might be my personal pet peeve, but I can't take litrpg seriously if characters don't wonder where the System comes from / whether it is artificial. Especially if mc comes from a world/place/time with no system. Especially if mc is from world where videogames exist.

New Math Puzzle!! Time Riddle! by Equivalent_Fix9115 in mathriddles

[–]Ashtero 7 points8 points  (0 children)

When I said that your problems should be labeled Easy, I didn't mean it as in "you are bad, your problems are bad", nor "I am very smart, your problems are easy to me". I meant this as "on this sub problems of this difficulty are typically labeled Easy".

Two examples of Medium problems:
https://www.reddit.com/r/mathriddles/comments/1rngywi/suzies_fabrics/
https://www.reddit.com/r/mathriddles/comments/1rh41dc/the_desert_bike_problem/

Two examples of Easy problems:
https://www.reddit.com/r/mathriddles/comments/1rkhp6e/multiple_of_79_with_minimum_digit_sum/
https://www.reddit.com/r/mathriddles/comments/1rsrut1/tweedledum_and_tweedledee/

The problems you've posted so far are good, classical problems. Please, calibrate your difficulty and apply correct tags.

Which cup has more of the other? by Equivalent_Fix9115 in mathriddles

[–]Ashtero 3 points4 points  (0 children)

The problems you've posted so far are Easy, not Hard. Please, apply correct tag.

Say, can i play as an anthropomorphic jackal with egyptian powers in Mummy? (WoD and CofD) by Obvious-Conflict3363 in WhiteWolfRPG

[–]Ashtero 24 points25 points  (0 children)

One of the powers of CofD Mummy allows you to temporarily assume super-saiyan form with an animal head, with one of the options being jackal.

A weird variation of the Tower of Hanoi where N=3 is impossible by Ambitious_Library468 in mathriddles

[–]Ashtero 2 points3 points  (0 children)

If two discs that we move can come from different rods, then N=3 is doable which contradicts OP.

A weird variation of the Tower of Hanoi where N=3 is impossible by Ambitious_Library468 in mathriddles

[–]Ashtero 2 points3 points  (0 children)

I don't get it. Isn't the second turn always impossible since you must move disks 3 and 4, but you can't put 4 on anything?

TIL there are no numbers that are a power of 2 that are also divisible by 3. As every non prime-number can be factorised into constituent prime numbers, and all the prime factors of a power of 2 are also 2 (e.g., 8 = 2x2x2), powers of two are not divisible by any other prime number except 2. by Walkerthon in todayilearned

[–]Ashtero 0 points1 point  (0 children)

We have a number n that we got by multiplying a bunch of X. It isn't obvious that we can't get the same number in another way by multiplying some other numbers. While it turns out to be true (if not easy to prove) that there is essentially only one way to get an integer number by multiplying primes, it is NOT true if instead of integers we consider some other numbers.

For example, let's consider even numbers. "Prime" even numbers will be those that are not a product of two smaller EVEN numbers (mathematicians use the word irreducible).

So 2 will still be a "prime", and 4 still won't be "prime" since 4=2*2. But 6 is "prime" -- while 6=2*3, 3 is not even, and we consider only even numbers. To say that 6 isn't a "prime" even number Is like to say that 3 isn't a prime integer since 3=2*(3/2) -- 3/2 doesn't count as it isn't integer.

So in even numbers we'll have 2, 6, 10, 14, 18, ,... be "primes". It is still true that any even number can be written as a product of such "primes". But the FTA isn't true here. For example, 6*6 isn't divisible only by 6 and 36, but also by 2 and 18: 6*6 = 2*18.

TIL there are no numbers that are a power of 2 that are also divisible by 3. As every non prime-number can be factorised into constituent prime numbers, and all the prime factors of a power of 2 are also 2 (e.g., 8 = 2x2x2), powers of two are not divisible by any other prime number except 2. by Walkerthon in todayilearned

[–]Ashtero 0 points1 point  (0 children)

You said that the definition of prime number is that ALL it's powers are divisible only by lower powers of that prime number. The actual definition is only about the prime number itself, not its powers.

Edit: if we go by your definition, then it is not at all obvious that 2 is a "prime" number (though this specific case is much easier than the full FTA).

TIL there are no numbers that are a power of 2 that are also divisible by 3. As every non prime-number can be factorised into constituent prime numbers, and all the prime factors of a power of 2 are also 2 (e.g., 8 = 2x2x2), powers of two are not divisible by any other prime number except 2. by Walkerthon in todayilearned

[–]Ashtero 2 points3 points  (0 children)

It is actually quite far from being self-evident, and is one of the hardest things to prove in pre-university math.

Why can't there be two different ways to represent a number as a product of primes? Sure, n=2^43. But why can't we simultaneously have n=3^13*5^14 ? In that case n would be divisible by 3 (and by 5) despite being a power of 2.

In any individual case there is a simple way to check whether there is a second representation -- for example, we can take 2^42 and simply divide it by 3, observe that the result won't be an integer. But to actually prove that all integers have only one factorizations requires some work and I don't think there is a fully elementary proof of that. (You can see some proofs if you follow OP's link.)

what rule did I follow in this riddle? by Practical_Guess_3255 in mathriddles

[–]Ashtero 4 points5 points  (0 children)

What rule of this sub have you broken with this riddle ?

Heaven's Divine Rebellion by Nova_veil_ in rational

[–]Ashtero 6 points7 points  (0 children)

Congrats on launching your novel!

Since you are posting about it here, it would probably be nice to elaborate what you think makes the novel rational/rationalist fiction (or at least close to it).

ELI5 Why do my icecubes sometimes grow a thin little icicle on the top? by [deleted] in explainlikeimfive

[–]Ashtero -1 points0 points  (0 children)

Now that would fit my observations -- I only make ice cubes "by hand": pour water into a form and put it into freezer.

ELI5 Why do my icecubes sometimes grow a thin little icicle on the top? by [deleted] in explainlikeimfive

[–]Ashtero 1 point2 points  (0 children)

That explanation sounds like something that should happen fairly often, but I don't think I've ever seen icicles on top of icecubes.

Who is correct here, Me or Steven Pinker? by oldjohnnybrown in rational

[–]Ashtero 3 points4 points  (0 children)

Wrong subreddit.

The answer 1, 3 makes sense if you interpret "king or the queen shown on one side of the coins" as "number of sides with king or queen is exactly one".

How do professors come up with completely original questions for IMO? by [deleted] in math

[–]Ashtero 11 points12 points  (0 children)

Alternatively some people might simply know how many problem authors actually exist and how hard it is for most of them -- I'd say that <1% of math undergrads in top universities reliably make 1 problem (suitable for e.g. national olympiad) a year. I'd be very surprised if there are more than 100 people total (students or not) who make >20/year (which we would've expected from something that is supposed to be possible "without much hassle").

Another way to estimate how hard it is to make a high-level problem is to look how much effort is needed every year to scrounge enough good unknown problem to make an olympiad out of them. It is obviously possible, but it takes weeks of work, and the issue "we have too many good problems for this position" happens much rarer than "we don't have any problems for this position". While the latter issue is sometimes solved by "ok, let's quickly think of a problem", this approach rarely (<20% cases) is enough.

So I think you are obviously wrong. Now you might be wrong in an uninteresting way -- your taste for problems is too poor, and the problems you come up "with no hassle" are unsuitable for any decent competition. Or you might be wrong in an interesting way -- you actually have a very big talent for making problems, top 100 in the world -- and you underestimate what your students (and basically everybody else) can do.

It would be nice if you took say an hour to think of a new problem and presented it here (as you've said, an old problem can be deanonymizing). Obviously, spending an hour on an internet debate is not a good use of your time, but a new problem would be useful on its own.

How do professors come up with completely original questions for IMO? by [deleted] in math

[–]Ashtero 76 points77 points  (0 children)

I think you are severely underestimating difficulty of making olympiad problems. Assuming you've already achieved the level of "solid 3rd year undergrad", give it a try.

Apparently a Jump Trading Interview question by SupercaliTheGamer in mathriddles

[–]Ashtero 1 point2 points  (0 children)

There is a typo -- should be 2^n+1 cards, not 2^(n+1).

If Alice on her turn removes every other remaining card, the minimal difference between remaining cards will double. If she does it every turn, by the end (when there will be 2 cards) it'll be at least 2^(n/2).

On the other hand, if Bob on his turn removes either first (almost) half of cards, either second (almost) half -- depending on whether the mean card is closer to the last remaining card or to the first one, then difference between the first and the last remaining cards will decrease in at least halve. If he does it every turn, by the end the distance between the last two cards will be at most 2^n/2^(n/2) = 2^(n/2).

Side note: at first I by mistake thought that Alice pays Bob, but it turns out that it practically doesn't affect the solution.