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Nice result.Quote:
Originally Posted by desiresjab
I've seen the technique used as a way of showing that a repeating decimal can be represented as a ratio of integers. This shows that such numbers are "rational". However, I did not realize that this implied that any prime has a multiple where the product is all nines. Now that you point it out, it makes sense that this should be the case.
I will give the forum three days to work on a problem that took me longer than that to solve. If more time is requested, I will give it gladly.
* * * * *
If I summed all the digits of the decimal representation of 44444444, and called that number C, then summed all the digits of that number, what would the resulting number D be, exactly.
It looks like this: AA=B, the sum of the digits of B=C, and the sum of the digits of C=D. Find D.
Can anyone do this problem? Give it a try.
I wouldn't be able to get beyond the neat way you present the problem, desiresjab.Quote:
Originally Posted by desiresjab
Well, this problem is no slouch. I believe I found it in an old math Olympiad exam. It was a bonus question for the brightest teenagers on earth. That is kinda disgusting, isn't it? Although I have had possession of the problem for years, I did not look at it much until I began to collect some tools. Then it took a long time of laying it down for long spells and forgetting all about it. A problem I want to solve but have no way to approach can drive me crazy if it stays in my mind. I just loved the shape of this problem and what it was asking for looked so impossible.Quote:
Originally Posted by Dreamwoven
What solving stuff like this comes down to is having the tools. There are always properties, laws and techniques which are related somehow and can be used to excavate the answer. If you have no idea about the existence of some property which will lead you right to the answer you are in for a long haul, probably an impossible one. Only cats like Leibniz could solve such a monster unprepared, independently rediscovering such laws and properties as are needed along the way.
I will show how this one is done soon. Right now I have personal obligations begging to be resolved around the house.
Sum the digits of the decimal representation of 44444444, then sum the digits of the new number. The form of the problem looks like this:
AA=B, the sum of B's digits equals C, the sum of C's digits equals D. Find D, exactly.
To find the number of digits in the decimal representation of 44444444, in other words how long the number is, the algebraic technique is to take the log of and add 1, while chopping whatever remains behind the decimal, leaving us with a whole number:
AA, like so:
↓{1+log AA}↓=↓{1+A log A}↓=16211, in the case of
44444444.
The actual calculations are as follows:
4444(log 4444)=16210.707879.... After we chop the decimal, which is irrational and goes on forever without pattern, and add 1 we know the huge number AA has 16211 digits. This number is far larger than the number of atomic particles in the universe, which is only in the neighborhood of 10120 particles, tops.
How can we sum the digits of B when we do not even know the number? We detour creatively, by supposing every one of those 16211 digits to be a 9, so that when we sum them they will equal 9(16211)=145899.
145899, then, is the upper limit for C, it can be no larger. But we are allowed to pretend again. We pretend that all six digits of C are equal to 9.
6X9=54. Ah, do you know what that is? It is an upper limit for D, the number we are after.
Suddenly, we are somewhere. One can almost smell solutions, but how do we get there? The answer lies again in observation and technique, in that order.
When any huge numbers are multipled, it is always easy for us to ascertain what the last digit is. In the case of 4444 times itself, no matter how many times successively we perform the operation, the last digit is always 4 or 6. Even powers of 4444 end in 6, and odd powers end in 4. Of course 4444 as an exponent is an even power, so 44444444 ends in a 6, and when divided by 10 would leave a remainder of 6.
We must detour now for some pretty facts, lest we arrive at our final destination with our route still shrouded in mystery.
* * * * *
Any time you sum the digits of a number X, that sum S will remain congruent to X (mod 9) through successive summing operations. This only happens with 9 because we use base 10. In base 8, 7 would have this same property through successive summing operations. Any base. (The above preservation property is also true for 3 in base 10).
Notice that once we ascertained the number of digits of 44444444, summing the digits successively is the only operation we have performed.
Successive powers under a modulus (any particular divisor ) always bend back and reapeat themselves in a cycle. This is called a power residue cycle, a cyclotomic number, to throw in a fancy term dangerously. They are certainly cyclical, but I do not know if that makes them cyclotomic. The host confesses.
A Modulus does not allow any number in its system to be as large as itself. The modulus is king. Larger numbers are bent back by division until only a remainder is left. Any whole number, when divided by 10, for instance, will leave one of ten remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. That's it. The modulus can systematically reduce any giant number to one smaller than itself. And as far as the king is concerned, any larger number than itself which it can divide is equal to zero. 30=0 (mod 10).
Let's look at the power residue set of 4444, (mod 9). It is okay to begin with the zero power of 4444, which we know is always 1 for any number, and 1 will always be the first number in the power resudue set of any number.
44440=1 (mod 9)
44441=16=7 (mod 9)
44442=19749136=40=4 (mod 9)
44443=87765160384=55=1 (mod 9)
...
...
...
44444444=1 (mod 3)
9(493)=4437, means 4444=7 (mod 9)
44444444=7 (mod 9)
At its 4444th power, 4444 is congruent to 7 (mod 9) and 1 (mod 3).
In fact, the power residue set for this number (mod 3) is
{1, 1, 1, 1, 1....}, and on any power our huge exponential number only equals 1 (mod 3).
The pattern listed above vertically for (mod 9) power residues has produced 1 again at power 3, so we know the complete cycle, which goes 1, 7, 4, repeating every three powers until the highest power is reached. Also notice that to take the congruency of a number with respect to 9 in our base 10 system, we only need to sum its digits then take the remainder when divided by 9.
The more convenient way for calculation is to begin the pattern on the first power so that the power residue set cycles like this from power 1:
{7, 4, 1, 7, 4, 1...}
* * * * *
Gathering what we know so far:
Our target number is less than or equal to 54 and is congruent to 7 (mod 9).
That becomes a small set.
{7, 16, 25, 34, 43, 52}
We have gone from fifty-four possiblities to six.
Now, as was the case with 9, the value, the magnitude, of a number is the same as the sum of its digits (mod 3), so we are further able to say that the correct answer must be congruent to 1 (mod 3). Only three numbers now qualify.
{7, 16, 34}
Now the judging becomes more demanding, more appropriately the search for a tool or a property to distinguish Miss America out of the three.
We know that B ends in a 6. In symbols B=6 (mod 10). But how does this help us? 10 cannot play the same trick that 3 and 9 did, for the preservation of its congruency does not happen. (It would happen on 10 only for base 11, where the factor 5 would also exhibit the property). Ah, but maybe 6 can play the trick. Actually, it cannot help us that way, either.
The simple observation that works is that any number which leaves a remainder of 1 when when divided by 3 and a remainder of 7 when divided by 9, has to leave a reaminder of 4 when divided by 6. But the English of that is so long and messy. Check this out.
If D=1 (mod 3) and= 7 (mod 9), then D=4 (mod 6).
Only one number from the last set qualifies:
{16} is the answer. D=16
One can only assume several of the interior digital places of C are 0 and 1.
Since 16 is smaller than expected, let me recheck the steps. The mistake, if any, is probably in following that (7, 4, 1) cycle correctly to the last power.
I may be back.
I think I can proclaim it correct, since I validated my stepping on the power residue set of 7, 4, 1. At the 4443rd power the cycle is on 1, which puts it on 7 for the 4444th power.
Since we know C is less than 145899 but still a six digit number, I envision a number built along the following lines:
100456. It makes perfect sense now.
I had to scratch the next post out and replace it with this. Everything is all right. I know the answer is correct.
To see if I understand, you have done the following:
1) Calculated the number of digits in 44444444 using a logarithm. You found there to be 16211 digits.
2) Found an upper bound on the sum of those digits by assuming they are all 9's, the highest value each digit could be. The max sum is 145899 which would be the maximum value of C. We know there are 6 digits in that sum.
3) We want the sum of the digits of C and so can find an upper bound on this as done in (2) by setting all the digits to 9. That max sum is 54.
4) You note that 44444444 (mod 9) = 7 and the sum of the digits of that number is also 7 (mod 9). This reduces the possibilities for D to a number in this set: {7, 16, 25, 34, 43, 52}
5) You note that 44444444 (mod 3) = 1. This reduces the possible value of D to one of {7, 16, 34}.
6) I am still trying to think through the last part, but I will leave this till later.
Quote:
Originally Posted by YesNo
Your confusion at step six may be because indeed I have overlooked something. 16 and 34 both meet all three qualifications, i.e. =7 (mod 9), =1 (mod 3), and =4 (mod 6).
So really, our set of prospects still has two members left:
{16, 34}.
Now I have to think of something else.
Ha! ha! The laugh is on me. But I think I have the fix through a different method on step six of Yes/No's listing of my steps.
The idea is to go to base 11 and cast out 10's. We do not actually need to go to base 11, for that would involve adding precisely one new symbol (usually designated simply as a), so let's just say we did and not, because we know 44444444 is the same value whether it is expressed in base 10 or base 11, and so would also yield the same value when 10's were cast out, whether that identical value used identical digits in the same order in the two systems or not. This way we will know how the sum of our number (mod 10) behaves and we are enabled to preserve our value through successive operations of summing the digits. These operations can be imaginary too, since they preserve through. When we cast out 10's, just as we cast out 9's in base 10, we are left with 6, which has identical real value in base 10 or base 11.
By this reasoning, this time I believe infallible, 16 is still the correct answer
A Chinese parcel delivery worker just made a significant discovery concerning Carmichael numbers. Those are an important class of composite number that is hard to factor and often fools primality tests for Fermat primes. One of the beauties of number theory is that discoveries like this one are often made without the use of advanced tools.
The casting out 10's seems promising. I haven't tried it but I expect one should be able to distinguish which of those three numbers is the correct answer.
Do you have a link to the information about the discovery concerning Carmichael numbers? I was in Madison, Wisconsin, yesterday and stopped by some of the many used book stores in that university town and found Carmichael's "The Theory of Numbers" and "Diophantine Analysis". I was planning on going through that book making a Jupyter notebook out of some of the problems I found interesting.
The links I found were distinctly uninteresting. They said less than I have. A Chinese friend told me about it. "Chinese amateur Carnichael numbers", worked as a search phrase.Quote:
Originally Posted by YesNo
I believe the casting out of 10's does work. I am very happy with it because I have never seen that used before. It will be embarrassing if my logic is wrong.
The book I am after is called 52! I had a book going on a similar subject when I heard about this one being published. I want to see if he did a better job than I think I could have done. I still may finish mine, if I do not think too much of his effort. I could order it online, but I do not make any financial transactions over the internet. If you find this one score it for me and I will pay cost plus shipping it to me.
I searched for "52!" but I only came up with links to "52 factorial". Do you know the author or something more about the book?
I'm not familiar with casting out 10's. If I get some time I will see if the idea would work. However, you might want to put in the details of the proof.
It doesn't hurt to publish your own book even if you liked the other one. However, it is good to have a list of references and that book may be one of them.
Michael Wayne Cottle write 52!Quote:
Originally Posted by YesNo
I was not familiar with casting out 10's, either. It was just an idea that came to me.
You can find the book on Amazon. The kindle edition is 99 cents.Quote:
Originally Posted by desiresjab
He sounds like an interesting character.
Tell me more about casting out 10's. I don't really follow it.
Is there a certain part of the demonstartion I might explain more clearly, Yes/No?
The biggest leap of faith comes in going to base 11 so we can see what happens to the sums (mod 10), without actually ever going there. I reason that a number expressed in one base versus another does not have more factors of 10. The digit 6 is congruent to this set
(mod 10) {6, 16, 26, 36...} and this one (mod 11) {6, 17, 28, 39...}
It seems to me that for a number A to leave a remainder smaller than itself equal to 6 in (mod 11), that remainder can only be 6, though the original number in base 11 might have ended in a digit other than 6. In fact, the number may end with the digit 5 in base 11, like 105 does. The reasoning for theoretically going to base 11 is to preserve the nature of (mod 10) sums over the operation of summing the digits of the powered number several times, as they are for 9 in base 10. I believe we can only guarantee this with a modulus which is one less than the base.
So when I sum all these digits in (base 11), (mod 10), I should get the real last digit. I already know what B is (mod 10), but without going to base 11, I could not say anything about the last digit of D, however, because the result of the summing operation would not be preserved. The sum of a number's digit ought to stay congruent to the same set (mod 10).
I have never consulted anyone because I don't like to do that, since I am trying to solve problems myself. I know a math phd who I have played music with. At this point I would not mind giving him a call, to see if my reasoning has been correct. He will quickly see other ways to do it, so I will have to keep him on track. Even if I got the right answer, I want to be sure my reasoning was not faulty somewhere.
Because the problems I choose are hard for me, that makes them fun even when they are torturous. I always learn a lot and correct myself a lot.
So there are three candidates to consider, {7, 16, 34}, in base 10. We got that set by considering 44444444 = 7 (mod 9). Wouldn't we have to rewrite 44444444 in base 11 and find out what the digit was (probably not 7) modulo ten in base 11?
My suspicion is that converting to a new base might not help with the solution, but I don't know.
It is easy to verify that B ends in 6, base 10. Simply by multiplying single 4's together and watching the last digit. It only alternates between 4 on odd powers and 6 on even powers. If one doubted that technique, one could go here http://www.javascripter.net/math/cal...calculator.htm and calculate the number directly and observe all 16211 digits--a number I was relieved to see matched my own.
Then the other day I went over to another specialized calculator that was supposed to calculate the sums of digits. I pasted in the digits of our AA=↓B↓=C and it gave me back a measley sum of seventy-six thousand and something. This was only a five digit number, and so I felt it had to be wrong. I wrote the publishers of the calculator and told them so. Now I begin to waffle on my own judgement there.
C represented a maximum value of the sum of B's digits. Maybe it was six digits long in the case of that maximum, but only five digits in reality. Then I would have gotten my maximum for D from these over estimates, which is perfectly fine, since all I was looking for at that point was a maximum.
* * * * *
I was so convinced the sums of digits calculator was wrong that I did not even bother to add up the number it gave as C. Had I done this, I would have noticed that the sum of the digits of 72,601=16, like I did just minutes ago. It was seventy-two thousand and something, not 76 and something. I did not notice the details. All I thought I saw was the wrong answer for C sitting there.
This means I must have arrived at the right answer despite some faulty conclusions. I was protected by the part of my reasoning which was correct--that D was a maximum of 54 and conguent to 7 (mod 9) and 1 (mod 3). It did not matter at all that I thought C was a six digit number instead of a five digit one. It did not interfere with getting the answer.
It seems the answer is settled, then, by brute force and somewhat inadvertantly. It also seems highly unlikely that I would have come up with this particular answer through erroneous reasoning. All my work pointed to this answer.
My base 11 speculations are open for discussion. They twist the brain like a sponge.
* * * * *
I don't know if I mentioned this, but since 4447 is a prime, I thought there might be some means of working backwards through Fermat's little theorem to 4444. It was a fruitless avenue, but I may have only taken a wrong turn.
* * * * *
The key piece of reasoning is in realizing that the congruence class of the sums of digits is preserved mod 9 across multiple operations. The other was understanding power residue sets and their behavior.
I am satisfied with the discussion down to the set {7, 16, 34}. I had already eliminated 7, but I cannot remember how right at this moment, so it can stay in the discussion.
My speculation about base 11 was correct!
The letter A in the following has nothing to do with the letter A in AA, but is the letter of the alphabet used to represent 10 in base 11.
The base 11 digits of a number X will always add up (mod 10) to something with the same congruency (mod 10) as the decimal representation of X.
Base
10----11
96----88
196--169
296--24A
Whatever those digits base 11 add up to, we know they will be congruent (mod 10) to whatever the base 10 representation is congruent to. And we further know that the congruence class is preserved across the operation of summing the digits of the base 11 representation. This is how we know with certainty that the last digit of D is 6 in decimal representation, just as it is the last digit of B. I believe that is all of it.
The former topic feels so wrapped up that I feel like another problem to untangle how God thinks is due. I believe the following was from a prep test for math Olympiad.
Prove that (2m)! (3n)! is always an integer.
..................(m!)2 (n!)3
(2m)! (3n)! The bottom factors easily, not the top.
m! m! n! n! n!
m! 2mΠm+1 n! 3nΠn+1 after cancellation, this becomes:
m! m! n! n! n!
2mΠm+1 3nΠn+1
m! n! n!
2(m) 2(m-1)...(m+1) 3(n) 3(n-1)....2(n) 2(n-1)...(n+1)
...m...(m-1)...(1)--------(n)....(n-1)......(n)...(n-1)...(1)
All the terms above are factors in both numerator and denominator. I had to use to dots and dashes to make them line up properly for illustrative purposes.
We see that any prime up to M in the denominator will have a double in the numerator, and will so be cancelled. Any prime up to N in the denominator has both a double and a triple in the set of numbers n+1 through 3n, and so both powers of N! are cancelled, leaving a whole number in the numerator and 1 in the denominator for both M and N.
This was the first time I ever figured out how to factor a factorial notatively.
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
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.
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?
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.
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.
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?
I know that thing is possible. It just hurts my head like h3ll, though.
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.
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.
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.Quote:
Originally Posted by YesNo
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.
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.
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.
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
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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.
Oh, boy. All our questions and more are answered in this little paper. The authors give examples of sums of squares with 1, 2, 3, 4, 6 (Can't remember if 5 was there) unique expressions, or factorizations, as these authors more properly call them.
A thorn in the path to watch for is their not counting 02 in their method of calculation. This means the standard formula (which I found elsewhere) will give different values. One only needs to subtract 1 from the total number of expressions.
The standard forumla for the number of expressions of any natural number to any power is:
rk(n)=2[d1(n)-d3(n)].
Now n is the sum of squares itself, k is the power, which may be higher than 2, and d1 and d3 signify the number of actual factors or divisors of n, prime or otherwise, which are respectively congruent to 1 or 3 (mod 4). A congruency of 2 or 0 (mod 4) is not germane, and not included in the calculations.
http://www.rowan.edu/colleges/csm/de...Submission.pdf
Now as far as the two professor problem goes, I have made a terrible, amateurish oversight. I always confess to my oafish ignorance and embarrassing oversights, especially the real whoppers, because they make such good copy.
Essentially, words two and three of the problem nail down a simple constraint I flat overlooked in my zeal to forge ahead. Our ignorance always forces learning upon us when we persist. I cannot be emabarrassed for what I have learned, but only for what I used to not know, like yesterday.
We learned a lot, and the problem will be quite different when we return to it tomorrow--with some elements ejected from the set of prospects.
This article made me realize that whenever one is talking about sums of squares one should think Pythagorean theorem and circles.Quote:
Originally Posted by desiresjab
Quite true. Remember that is what the Martinson article did.Quote:
Originally Posted by YesNo
The two professor problem is too hard for me unless I find another line of attack. Plus, it is not a fun problem for me, but has become a minor obsession anyway. For recreation I have to do beautiful problems. At least they are beautiful to me. I wonder if anyone agrees with me over what "looks" good in mathematics. Here is one.
When is xpq divisible by (xp)q? Prove.
The last problem is very easy of course. Let pq=a, let pq=b.
From basic algebra:
xa=xa-b.
xb
Problems like this are good for refreshing yourself on technique.
The answer is for all x except 0.