Challenge Math Challenge - January 2019

[fresh_42]

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?

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

 

[scottdave]

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

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.

 

[fresh_42]

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.

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.

 

[PeroK]

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.

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

 

[fresh_42]

... but what is meant is "an extra coconut upon division by three"

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

 

[PeroK]

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

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!

 

[scottdave]

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!

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.

 

[fresh_42]

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.

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 $(26+7,17+7,11+7,3)=(33,24,18,3)$.

 

[member 587159]

9 (b):

Spoiler

Answer is no:

Take your favorite non-commutative ring $R$ and the trivial ring $S = \{0\}$.

Then the map $\phi: R \to S: r \mapsto 0$ is clearly a ring epimorphism and $\phi^{-1}(Z(S)) = \phi^{-1}(0) = R \neq Z(R)$.

 

[Periwinkle]

8. Let $G$ be a group generated by $\sigma,\varepsilon,\delta$ with $\sigma^7=\varepsilon^{11}=\delta^{13}=1$.

• Show that there is no transitive operation of $G$ on a set with $8$ elements.
• Is there are group $G$ with the above properties, that operates transitively on a set with $12$ elements?

There is no transitive operation of $G$ on a set with $8$ elements​

$G$ is a transformation group, so a permutation of the eight-element set corresponds to each element of the group by a homeomorphism. Each permutation can be produced as a product of distinct cyclic permutations in one way. Since $\varepsilon^{11}=\delta^{13}=1$, consequently, identity permutation corresponds to both $\varepsilon$ and $\delta$ elements.

For the $\tau(\sigma)$ homeomorphic equivalent of the $\sigma$ element, $\tau(\sigma)^7$ is identical to identity permutation. Therefore, this permutation is identity permutation or $(a_1,a_2,a_3,a_4,a_5,a_6,a_7)(a_8)$.

The transformation group is transitive if any $x$ element of the eight-element set can be transformed to any $y$ element using permutation corresponding to an element of the $G$ group. However, the $a_1, a_2, \dots, a_7$ elements in the $a_8$ element is not transformed by any power of the permutation corresponding to $\sigma$.

Group $G$ with the above properties, that operates transitively on a set with $12$ elements​

$\sigma = (a_1,a_2,a_3,a_4,a_5,a_6,a_7)(a_8)(a_9),(a_{10}),(a_{11}),(a_{12})$
$\epsilon = (a_1)(a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10},a_{11},a_{12})$
$\delta$ is the identity permutation

$a_1$ is transformed to $a_2$ by $\sigma$. With the power of $\epsilon$ transformation, we can further transform into any element other than $a_1$.

$a_2$ is transformed to $a_1$ by $\sigma^6$. However, the element $a_2$ can be obtained from any element other than $a_1$ by the corresponding power of transformation $\epsilon$, so that the $a_1$ element also occurs as a transformed image of any element other than $a_1$.

 

[fresh_42]

There is no transitive operation of $G$ on a set with $8$ elements​

$G$ is a transformation group, so a permutation of the eight-element set corresponds to each element of the group by a homeomorphism. Each permutation can be produced as a product of distinct cyclic permutations in one way. Since $\varepsilon^{11}=\delta^{13}=1$, consequently, identity permutation corresponds to both $\varepsilon$ and $\delta$ elements.

For the $\tau(\sigma)$ homeomorphic equivalent of the $\sigma$ element, $\tau(\sigma)^7$ is identical to identity permutation. Therefore, this permutation is identity permutation or $(a_1,a_2,a_3,a_4,a_5,a_6,a_7)(a_8)$.

The transformation group is transitive if any $x$ element of the eight-element set can be transformed to any $y$ element using permutation corresponding to an element of the $G$ group. However, the $a_1, a_2, \dots, a_7$ elements in the $a_8$ element is not transformed by any power of the permutation corresponding to $\sigma$.

Group $G$ with the above properties, that operates transitively on a set with $12$ elements​

$\sigma = (a_1,a_2,a_3,a_4,a_5,a_6,a_7)(a_8)(a_9),(a_{10}),(a_{11}),(a_{12})$
$\epsilon = (a_1)(a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10},a_{11},a_{12})$
$\delta$ is the identity permutation

$a_1$ is transformed to $a_2$ by $\sigma$. With the power of $\epsilon$ transformation, we can further transform into any element other than $a_1$.

$a_2$ is transformed to $a_1$ by $\sigma^6$. However, the element $a_2$ can be obtained from any element other than $a_1$ by the corresponding power of transformation $\epsilon$, so that the $a_1$ element also occurs as a transformed image of any element other than $a_1$.

Correct. Let me add my (identical) version, as I think it becomes more transparent with less indices:

Assume $G$ operates transitively on $M=\{\,1,2,\ldots ,8 \,\}$ via $\varphi\, : \,G\longrightarrow S_8\,.$ As the order of $\varphi(\varepsilon)$ is a common divisor of $11$ and $|S_8|=8!$, and both numbers are coprime, we thus get $\varphi = 1\,.$
The same argument applies to $\varphi(\delta)$ hence $\varphi(\sigma)$ generates $\varphi(G)$, which is a cyclic group of order $1$ or $7$.
By the orbit-stabilizer theorem, for a transitive operation we would have $8\,|\, |\varphi(\sigma)|=|\varphi(G)|\in \{\,1,7\,\}$ which is impossible.

Let $\sigma = (1\,2\,3\,4\,5\,6\,7)$ and $\varepsilon =(2\,3\,4\,5\,6\,7\,8\,9\,10\,11\,12)\,.$
Both cycles generate a subgroup $H \leq S_{12}$ which operates transitively on $M=\{\,1,2,\ldots 12\,\}\,.$ Now $(h,z).m := h.m$ is a transitive operation of
$$
G := H \times \mathbb{Z}/13\,\mathbb{Z}
$$
on $M$, too, and $G$ is generated by $(\sigma,0)\, , \,(\varepsilon,0)\, , \,(1,1+13\,\mathbb{Z})\,.$