FeaturedChallenge Math Challenge - January

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1. Jan 7, 2019

A slightly simpler way for problem 3: There is a (unknown) distance D between the trains. Let $v_t$ be the speed of the trains and $v_c$ the speed of the bicycle. $\\$ We have $30= \frac{D}{v_t-v_c}$ and $20=\frac{D}{v_t+v_c}$. The problem is to solve for $T=\frac{D}{v_t}$. The frequency will be the inverse of that. $\\$ Rewriting: $\frac{v_t}{D}-\frac{v_c}{D}=\frac{1}{30}$, and $\frac{v_t}{D}+\frac{v_c}{D}=\frac{1}{20}$. $\\$ Let $\frac{v_t}{D}=x$ and $\frac{v_c}{D}=y$. $\\$ This gives $x-y=\frac{1}{30}$ and $x+y=\frac{1}{20}$. $\\$ Adding these two equations: $2x=\frac{1}{12}$. $\\$ The result is $x=\frac{1}{T}=\frac{1}{24}$ (in frequency per minute), or $\frac{60 \, min/hr}{24 \, min}= 2.5$ trains each hour.

2. Jan 7, 2019

PeroK

... $f= \frac{1}{T} = \frac{v_t}{D} = \frac{1}{24}$

Even simpler!

3. Jan 7, 2019

Staff: Mentor

Imagine she rides one hour in one direction and one hour in the other. Then she meets three trains in the first hour and is overtaken by two in the second hour. So the frequency is thus $5$ trains per direction in $120$ minutes, i.e. every $24$ minutes a train.

4. Jan 8, 2019

Bilal Rajab Abbasi

Can we upload a picture of solution here?

5. Jan 8, 2019

Staff: Mentor

Please don't, type it out. Here's an info page on how to do this: https://www.physicsforums.com/help/latexhelp/
Pictures are almost always hard to read and often don't even have the correct orientation. Both forces the readers to download and edit it. I don't like being forced to do that.

6. Jan 9, 2019

Math_QED

Part 1 of (9)

$\phi: R \to S$ is a ring epimorphism. Define $I:= \phi^{-1}(J)$. It is well known that the inverse image of an ideal is an ideal, thus $I$ is an ideal.

Define $\psi: R/I \to S/J: [r] \mapsto [\phi(r)]$

This is well defined: If $r \in I$, then $\phi(r) \in J$.

Clearly, this is also a ring morphism.

For injectivity, assume $[\phi(r)] = 0$, then $\phi(r) \in J$, and $r \in \phi^{-1}(J) = I$, thus $[r] = 0$. The kernel is trivial and the map is injective.

Surjectivity follows immediately by surjectivity of $\phi$.

It follows that $\psi$ is an isomorphism, and thus $R/I \cong S/J$.

Last edited: Jan 9, 2019
7. Jan 9, 2019

Staff: Mentor

Correct.

You could have saved some lines by setting $\psi\,' := \pi \circ \phi$ with the canonical projection $\pi \, : \,S \twoheadrightarrow S/J$ and $I=\operatorname{ker}(\pi \phi)$ plus the isomorphism theorem.

I'll wait for the second part before I mark it as solved.

8. Jan 9, 2019

Math_QED

I'm studying exams, so I don't have much time to try now. It is sufficient to prove that $Z(S) = \phi(Z(R))$ and the inclusion $\supseteq$ is immediate by writing out the definitions. My first guess would be that the other inclusion can fail. A counterexample (if the statement isn't true) should be in non-commutative ring theory. Will come back to the problem once I got more time.

Maybe the first part of the question is relevant to this part?

9. Jan 9, 2019

Staff: Mentor

No.
Nobody said the rings were commutative!

I know that some distinguish between rings = commutative and pseudo-rings = not necessarily commutative, but a) I haven't learnt it this way, and b) do not see any advantage in it. A ring is an associative and distributive structure on an Abelian additive group. That's it. No mention of commutativity, nor unity.

10. Jan 9, 2019

scottdave

For future reference, I think that #2 (the pirates and coconuts) should be slightly reworded for clarity. It states that each pirate takes a third of the available coconuts, but then there is an odd number left, so he gives one away.

If we start with a quantity of coconuts, which is divisible by 3, let's say that is 3N coconuts, where N is a whole number. The first pirate takes a third of that (N coconuts), leaving 2N remaining, which is an even number. Perhaps the storyteller meant to say that it is not evenly divisible by 3, or something to that effect?

11. Jan 9, 2019

Staff: Mentor

If he leaves 2N coconuts, then there is no need to give away one for the monkeys!

12. Jan 9, 2019

scottdave

That's what I was thinking. But it appears that the problem intends something like this:
I am a pirate and wake up to steal some coconuts for myself and hide them. I'll take my "third", but I want to make sure the remaining pile is evenly divisible by 3. If the remaining pile is not, I will give away 1 or 2 (apparently only 1 was necessary) to make the remainder divisible by 3.
I expect to go back to sleep and wake up in the morning and we all get our "even share"

The next pirate wakes up with the same thinking. And the last pirate wakes up and does the same thing.

13. Jan 9, 2019

Staff: Mentor

Well, the given solution is the same as I intended and which was given where I took the riddle from. So instead to actively cheat the other pirates, the assumption seems to be passive, that every pirate wants to save his share before the others cheat on him.

14. Jan 10, 2019

PeroK

I think you're right. The problem says "an odd number", but what is meant is "an extra coconut upon division by three".

15. Jan 10, 2019

Staff: Mentor

No, because the pirate who took away his share, expects the rest to be divided by two, not three. No cheating intended here.

16. Jan 10, 2019

PeroK

Okay, honest pirates. Except, they all end up with different numbers of coconuts and they will all expect the other two to share the remaining coconuts. The whole funny business will be discovered and I suspect a knife fight would ensue in any case!

17. Jan 11, 2019 at 5:58 PM

scottdave

So if I wake up and take my third of the pile. If the pile is evenly divisible by 3 before I take any, then I can take away a whole number, and the remaining must be an even number, regardless of my intentions.

Let's say we do start with a multiple of 3. Let's represent that by 3n, where n is a natural number. Now when I take away n coconuts (my third), I leave 2n, which is an even number, not odd.

18. Jan 11, 2019 at 6:21 PM

Staff: Mentor

You're right. Had to be even $(52,34,22)$ and the next pirate to give away one for the monkey with 1 left for the monkeys at the end: $78-26=52\; , \;52-1-17=34;34-1-11=22=3 \cdot 7 +1$ makes $(26+7,17+7,11+7,3)=(33,24,18,3)$.