Cosmology

[YesNo]

The factorial problem seems to work as you described.

I also find it hard to trust the results of calculators especially when the numbers get large. Many things can go wrong: implementation, computer hardware, programming. I am currently using Python for other purposes and it looks like this should work well with number theory. You can get Python by installing the anaconda distribution at https://www.continuum.io/downloads. The software is free.

This should do multi-precision arithmetic. I also interface it with jupyter notebooks which comes with the anaconda distribution.

I checked the answer with Python to your previous problem and I also got 16 as you did. Here are copies of the results:

num = 4444**4444
total = sum( [ int(char) for char in str(num) ] )
72601
sum_of_total = sum( [ int(char) for char in str(total) ] )
16

I didn't know how to code the sum of digits and so I searched for a solution and used this one, that is, I didn't come up with it on my own: https://www.codecademy.com/en/forum_...453d00020186c8

[desiresjab]

Yes, sir, I believe we have the 4444 problem settled. The best part of it was what was learned along the way. After finishing a problem the next next problem is to find the next problem. One that is solvable is needed, but which will require considerable effort. One could work on the Goldbach conjecture, the twin prime conjecture or Brocard's problem, but would never have a reasonable chance of even making progress. But if one chooses an easier problem, it could lead to discoveries which might be of assistance on those unsolved questions later on.

I may throw in another Olympiad problem to hold us until a truly fascinating problem comes along.

[desiresjab]

The following is a type of problem I find extraordinarily difficult. Other people may see the answer fairly quickly. But I look at this thing and I am baffled where to even start. I have seen problems of this type which are even more brutal. I am sure there are number theoretic techniques to solve them, for I found this problem again in a prep test for math Olympiad. You see, I have a functional problem as a mthematician and a human being--if a technique looks ugly and cumbersome, I avoid it. I seem to be seeking the beautiful in mathematics. I post the following problem because it is so opposite to that, to me. It is quite brutal from my perspective. And since I have no techniques to solve it, it is a head-on brain against problem sort of deal. There is probably also a solution out of formal logic. Here the beast is:

Every man in a village knows instantly when another's wife is unfaithful, but never when his
own is. Each man is completely intelligent and knows that ev-
ery other man is. The law of the village demands that when
a man can PROVE that his wife has been unfaithful, he must
shoot her before sundown the same day. Every man is com-
pletely law-abiding. One day the mayor announces that there
is at least one unfaithful wife in the village. The mayor always tells the truth, and every man believes him. If in fact there are exactly forty unfaithful wives in the village (but that fact is not known to the men,) what will happen after the mayor's announcement?

[desiresjab]

Let this one stay up a for a while. I know some smart people would like to think about it. I may know the answer, but I am not exactly sure, either.

[desiresjab]

I think the question is rather poorly formed. It leaves a certain taste of ambiguity, especially what is in parentheses. That is why I am going to attempt to answer it already, and get it out of here. Then I will move on to a really really difficult one of this type which is well formed.

One of the villagers is probably the local mathematician. He asks each man in the village, including the mayor and himself, to count the number of wives they recall to have been unfaithful. They must only write down their own name and that number on their piece of bark. Then he collects each piece of bark and spreads them out for all to see. All forty men who recall only thirty-nine adulterous wives, must shoot their own before sundown. Even the mayor and the mathematician may end up shooting their wives. This works even if there are only forty-one men in the entire village.

Assuming that to be correct, it was not really that hard, I guess. Let us move on to a real monster, which I do not expect to be able solve at all.

[desiresjab]

Try this one on for size, folks. It is a real baffler. Yet it does have number information which can obviously be used to solve it. I have worked on this one before, and I see from my notes that my work ended in confusion and uncertainty. I will give it another shot, after trying to determine what I was up to before. Sometimes it is quite excruciating to reconstruct your own logic from forgotten work, especially if the logic happened to be wrong! I am on the line now. But to tell you the truth, I have no confidence at all on this one.



Two positive integers are chosen. The sum is revealed to logician A, and the sum of the
squares is revealed to logician B. Both A and B are given this information and the information
contained in this sentence. The conversation between A and B goes as follows: B starts
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` Now I can tell what they are.'

(a) What are the two numbers?

[desiresjab]

I know that thing is possible. It just hurts my head like h3ll, though.

[YesNo]

If all the men in the village are completely law-abiding, who is having sex with the unfaithful wives?

I might be missing something about the second problem. We know the sum of the two positive integers, call it S. Consider all the possibilities as A runs from 1 through S - 1 of the pairs A and S - A. Square each of these, A² and (S - A)², and see if the sum of those two squares equals the other known value. When it does then one has the two numbers.

[desiresjab]

I would approach this through two possible connections--the generalization of Fermat's theorem on the sums of two squares, in conjunction with the Pythagorean theorem. Only one might be necessary, or maybe neither. It could be wrong altogether, but this is where I would begin to sniff. It is necessary to know exactly what both professors learn from each answer of the other. One could perhaps walk upwards to the correct sum of squares this way. Brutal, but it could work. I am thinking perhaps the Pythagorean theorem can provide a shorcut to the answer, once one understands what is happening with each answer the professors give.

What you have to start with for this investigation is knowledge of just which numbers can be expressed as the sum of two squares. This is not too hard to remember: Those numbers exactly which either have no prime (4n+3) factors, or all prime (4n+3) factors in the prime factorization of the number are to an even power. That is, any number, as long as it prime 4n+3 factors are all to an even power. Either that or it has no such factor at all. Only such numbers can be expressed as the sums of two squares.

[YesNo]

I suppose we could shorten the brute force result with some algebra. Let the two unknown numbers be X and Y. We are given the sum of those numbers, X + Y, and the sum of the squares, X² + Y². As an example we can say the sum of the numbers is 6 and the sum of their squares is 26. We can change these parameters later. What are the numbers X and Y?

Since X + Y = 6, we know X = 6 - Y.

We can do the following transformation X² + Y² = (6 - Y)² + Y² = 36 - 12Y + Y² + Y² = 36 - 12Y + 2Y².

We are given that 36 - 12Y + 2Y² = 26, so we can subtract both sides of the equation by 26 and get the following quadratic equation: 10 - 12Y + 2Y² = 2(Y - 1)(Y - 5) = 0. There will be two solutions to this equation. If they are integers then we have the solutions we want. We can see that Y could be either 1 or 5. X would be the opposite.

[desiresjab]

I suppose we could shorten the brute force result with some algebra. Let the two unknown numbers be X and Y. We are given the sum of those numbers, X + Y, and the sum of the squares, X² + Y². As an example we can say the sum of the numbers is 6 and the sum of their squares is 26. We can change these parameters later. What are the numbers X and Y?

Since X + Y = 6, we know X = 6 - Y.

We can do the following transformation X² + Y² = (6 - Y)² + Y² = 36 - 12Y + Y² + Y² = 36 - 12Y + 2Y².

We are given that 36 - 12Y + 2Y² = 26, so we can subtract both sides of the equation by 26 and get the following quadratic equation: 10 - 12Y + 2Y² = 2(Y - 1)(Y - 5) = 0. There will be two solutions to this equation. If they are integers then we have the solutions we want. We can see that Y could be either 1 or 5. X would be the opposite.

Your algebra is beautifully done and succinct. It tells us if we plug in the right number we will get back the number we should. It just does not tell us if we have selected the right input number in the first place to make the two professors volley back and forth for six separate "I don't knows" before professor B has enough information to answer the question.

I learned a lot from looking at the following sequences. It is a list of the numbers which can be represented as the sums of two squares. Zero counts In symbols: 6²+0² is how one gets 36 as the sum of two squares, etc. The bottom rows are the sums of squares. The numbers above them are the number of representations for that number (how many different ways it can be represented as the sum of two squares.)

1, 1, 1, 1, 1, 1,, 1,, 1,,, 1,,, 1,,, 1,,, 1,,, 2,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1
1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45,


1,,, 2,,,, 1,,, 1,,,1,,,, 1,,, 1,,, 2,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 1,,, 2,,, 1,,, 1,,,1,,, 2
49, 50, 52, 53, 58, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 85, 89, 90, 97, 100


Something pops out immediatey: all the numbers representable in more than one way are divisible by 5.

If the sum of B's squares were anything less than 25, he would immediately know the decomposition, for there is only one composition of them. Both A and B realize this.

Let us assume the actual numbers are 5 and 0.

B knows A has a 3+4 or 5+0. A knows B has 13, 17 or 25.

As soon as B speaks, A knows B does not have a sum of 13, or 17, so he must have a 25. Since A's sum is 5, he knows the proper sum is 5+0.

* * * * *

Let us assume the actual numbers are 3 and 4.

B knows A has 3+4 or 5+0. A knows B has 49, 37, 29 or 25.

B says he doesn't know. This tells A B has 25.

A says he knows.

* * * * *

The same thing seems to happen at 50. The first number with enough complexity to keep the professors volleying might be 100. At least A does not know the answer when B speaks for the first time. Or perhaps it is the first number representable in three different ways, which I read was either 285 or 385, I think the latter.

The proper sum of squares must force the number of returns in the volley to seven responses total.

[desiresjab]

I wonder if each pair of responses eliminates one root. But that would be a cubic equation, since there are three pairs of call and response, and then B says he knows the answer. So far, as far as I can get is after B speaks, A also says he doesn't know, though I cannot determine what further information that gives B. That is for the sum of squares 100.

[YesNo]

I did not understand what the going back and forth was with A and B each saying they did not know the answer. Apparently they are not allowed to give each other the values they know. Each of them knows a limit on the possible answers without knowing each other's information.

[desiresjab]

My computer keeps freezing up. I lost a long post, but the solution was not in it. As soon as B speaks, A can eliminate all numbers but 50 and 100. But to B the set of possibilities in A's mind could be 50, 100, or 169. For, yes, 169 has two decompositions: 13²+0², and 12²+5².

And we see a doubly representable sum of squares does not have to be divisible by 5 after all. We have two intersecting sets sets {50, 100} and (100, 169), for 100 can be made from a sum of 10 or 14. Only A knows he is holding a sum X+Y=10, for instance, and not 14.

I will keep the posts relatively short to avoid freeze ups. To be continued as an editorial extension...right here

* * * *

[desiresjab]

The problem might be that the two sets {50, 100} and (100, 169) intersect. I have no idea whether non intersecting sets of doubly representable sums of squares are even possible. Their intersection causes problems for me. If non intersecting sets are possible, at this point I would have to guess the problem could be solved only with them, for I have trouble seeing where either one gains enough additional knowledge through the response when the sets intersect.