I had to scratch the next post out and replace it with this. Everything is all right. I know the answer is correct.
I had to scratch the next post out and replace it with this. Everything is all right. I know the answer is correct.
Last edited by desiresjab; 10-19-2016 at 04:43 AM.
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.
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.
Last edited by desiresjab; 10-19-2016 at 10:41 AM.
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
Last edited by desiresjab; 10-19-2016 at 05:36 PM.
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.
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.
Last edited by desiresjab; 10-21-2016 at 11:15 PM.
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!
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.
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.
Last edited by desiresjab; 10-23-2016 at 10:25 AM.
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.
Last edited by desiresjab; 10-24-2016 at 12:43 AM.
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.
Last edited by desiresjab; 10-24-2016 at 01:24 AM.
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