We are very lucky for Michael to share his passion about these. You know in many ways I think good teachers/trainers/motivators are really people with a passion for what they do. And perhaps good students/trainees/nominees are really people with a similar passion for what is done?
16:31 Now, is there any some prime number such that when you concatenate all the permutations the result is prime ? 37 would not work because 3773 and 7337 are not prime. 113 would not work because 113311131, 131113311 and others are not prime. Has someone found one working case?
well no 2-digit prime will work, since the concatenation will be divisible by 11. ( "abba" = 1001a + 110b = 11*(91a + 10b) ) meanwhile, 1 digit primes are unaffected so 2, 3, 5, 7 are trivial cases personally i doubt there exists such a prime that satisfies your condition, but i have no idea how to prove anything for longer primes (in a non tedious manner).
All apsolute primes are either repunits or permutation of aaa..aab where a,b are from {1,3,7,9}. This has been proven. For repunits concatenation is that prime itself so this case is trivially true. Now, let's look at n-digit numbers who have n-1 digits a and one digit b. It has been proven that for n>3 numbers of this form have to be divisible by 11088. But then the concatenation is divisible by 3 because sum of digits is divisible by 3. Let's prove that. Concatenation has n^2 digits, out of which n digits b and (n^2 - n) digits a, so the sum of all digits is nb + (n^2-n)a = n (b + an - a) which is divisible by 3 since since 3 | 11088 | n | n (b + an - a). Let's look at the case n=3, the number is of a form aab, there are 3!/2!=3 permutation, thus the number of digits of concatenation is 9, 6 a's and 3 b's, and the sum of digits is 6a+3b=3(2a+b).
Intriguing problem! Now I wrote some python code, factorized the resulting concatenated permutations of 3-digit prime numbers.. and i think i found the reason why such a prime can't exist in some cases, with a particular case being any 3-digit prime. The key thing I found is that if the number of permutations of the prime is a multiple of 3, (which happens for any given 3-digit prime, since, if it has two repeated digits, it has 3 permutations and, if it has no repeats it has 6 permutations, and also the repunit 111, of length 3, is not prime), then any concatenation of those permutations will have the digit sum of the prime repeated a multiple of 3 times, hence making the digital sum of the concatenated number a multiple of 3 (which makes the number a multiple of 3). @Alan-zf2tt Sorry but I don't understand the question. You mean to check 3-digit primes? The thing is, I happened to check for prime factors of particular given 3-digit primes and all of its concatenated permutations, via a python script using prime generating, prime factorization and permutation generating libraries. I figured it would be interesting to check for the primes with two repeated digits and the ones with no repeated digits, picked one of each and stumbled across the fact that all of the concatenated permutations were multiples of 3. Then I figured out the mathematical proof from above, which makes checking for any more of them not worth it. Also, in case you wanted to check if a number is prime, you only need to check for divisors
@@terak93 Is it easier to look at it from a modular math perspective? Computation is easy and google tells me there are 168 primes between 1 and 1000 I think - and may be wrong - that most employers would want a computational answer rather than a deep math proof. Unless, of course, the employer was Michael or a similarly motivated math prof. Usually these are referred to as "boss" or "assessor"? So .. either modular math on modulo p prime taken from a list of 168 options OR dividers taken from 168 list with zero remainders after division In either case there are finitely many permutations and finitely few "numerators" to be divided. And equally there is no point in dividing permutation candidate by a prime greater than the "numerator" note: where "numerator" denotes an acceptable permutation and "divisor" denotes an acceptable prime
Yes, but the ones he lists are definitely *among* all the permutations, and the existence of merely that set is enough to prove that one of them has to be composite.
Fourth with an edit and a correction... the edit @ 1:16 would it be an advantage to introduce notion of modulo math and relationship of numbers with the zero equivalence class in modulo p prime?
That is a class of toy problem because clearly it's not a property of the number itself but of the representation. For example, 73 is not an absolute prime in Binary but it's in base 8 but it's not in base 6. So we are really not learning anything about numbers but about dudes with 10 fingers. What is useful are the techniques for the proofs.
You can though apply the proof techniques in the video to any base to explore similar problems in those bases. So the proof techniques you’re learning here isn’t solely useful in base 10, modifications of it can be applied for other number systems.
This is a point that, in my opinion, has many answers. It is a bit like painting a still life. What value does it have when it has been done so many times before and to such a high standard of acclamation if we believe the acclaimers. In other words it is not the result that matters. It is the doing of math. There was a math researcher supposed to have something like 10 head and 21 legs. I forget the real numbers. Author had a French name, based in France doing, I think, cutting edge math research
Been waiting for a new video, I really appreciate what you do
We are very lucky for Michael to share his passion about these.
You know in many ways I think good teachers/trainers/motivators are really people with a passion for what they do.
And perhaps good students/trainees/nominees are really people with a similar passion for what is done?
The last part of the last line on the last board should be N + 6c(mod7) as 10100 - 6 is divisible by 7.
16:31 Now, is there any some prime number such that when you concatenate all the permutations the result is prime ?
37 would not work because 3773 and 7337 are not prime.
113 would not work because 113311131, 131113311 and others are not prime.
Has someone found one working case?
yes - I did but I lost it in the margin
well no 2-digit prime will work, since the concatenation will be divisible by 11. ( "abba" = 1001a + 110b = 11*(91a + 10b) )
meanwhile, 1 digit primes are unaffected so 2, 3, 5, 7 are trivial cases
personally i doubt there exists such a prime that satisfies your condition, but i have no idea how to prove anything for longer primes (in a non tedious manner).
All apsolute primes are either repunits or permutation of aaa..aab where a,b are from {1,3,7,9}. This has been proven. For repunits concatenation is that prime itself so this case is trivially true. Now, let's look at n-digit numbers who have n-1 digits a and one digit b. It has been proven that for n>3 numbers of this form have to be divisible by 11088. But then the concatenation is divisible by 3 because sum of digits is divisible by 3. Let's prove that. Concatenation has n^2 digits, out of which n digits b and (n^2 - n) digits a, so the sum of all digits is nb + (n^2-n)a = n (b + an - a) which is divisible by 3 since since 3 | 11088 | n | n (b + an - a). Let's look at the case n=3, the number is of a form aab, there are 3!/2!=3 permutation, thus the number of digits of concatenation is 9, 6 a's and 3 b's, and the sum of digits is 6a+3b=3(2a+b).
Intriguing problem!
Now I wrote some python code, factorized the resulting concatenated permutations of 3-digit prime numbers.. and i think i found the reason why such a prime can't exist in some cases, with a particular case being any 3-digit prime.
The key thing I found is that if the number of permutations of the prime is a multiple of 3, (which happens for any given 3-digit prime, since, if it has two repeated digits, it has 3 permutations and, if it has no repeats it has 6 permutations, and also the repunit 111, of length 3, is not prime), then any concatenation of those permutations will have the digit sum of the prime repeated a multiple of 3 times, hence making the digital sum of the concatenated number a multiple of 3 (which makes the number a multiple of 3).
@Alan-zf2tt Sorry but I don't understand the question. You mean to check 3-digit primes?
The thing is, I happened to check for prime factors of particular given 3-digit primes and all of its concatenated permutations, via a python script using prime generating, prime factorization and permutation generating libraries.
I figured it would be interesting to check for the primes with two repeated digits and the ones with no repeated digits, picked one of each and stumbled across the fact that all of the concatenated permutations were multiples of 3. Then I figured out the mathematical proof from above, which makes checking for any more of them not worth it.
Also, in case you wanted to check if a number is prime, you only need to check for divisors
@@terak93 Is it easier to look at it from a modular math perspective?
Computation is easy and google tells me there are 168 primes between 1 and 1000
I think - and may be wrong - that most employers would want a computational answer rather than a deep math proof. Unless, of course, the employer was Michael or a similarly motivated math prof. Usually these are referred to as "boss" or "assessor"?
So .. either modular math on modulo p prime taken from a list of 168 options OR dividers taken from 168 list with zero remainders after division
In either case there are finitely many permutations and finitely few "numerators" to be divided.
And equally there is no point in dividing permutation candidate by a prime greater than the "numerator"
note: where "numerator" denotes an acceptable permutation and "divisor" denotes an acceptable prime
I find these techniques really beautiful, thank you Professor!
7913 is congruent to 3, 9713 is congruent to 4, and 10100 is congruent to 6 (modulo 7).
Shows once again what a powerful problem-solving technique modular arithmetic is
3:21 Aren‘t there much more permutations of the set 1,3,7 and 9?
I suggest that you watch the full video
Yes, but the ones he lists are definitely *among* all the permutations, and the existence of merely that set is enough to prove that one of them has to be composite.
We need just 7 with different modulos (mod 7) out of a total of 4! = 24.
Last case I expected 6 (mod 7). As 10100 is 101 * 100, I get 3 * 2 (mod 7) = 6.
2:54 I was expecting a "year I was born in" joke.
The third number in the set (2:58) should be 9137 since it's congruent to 2(mod 7).
Please, could you prove that the shortest pipe that will get stuck in the corridor corner is the longest that passes through the corridor?
Fourth with an edit and a correction... the edit
@ 1:16 would it be an advantage to introduce notion of modulo math and relationship of numbers with the zero equivalence class in modulo p prime?
Absolute primes are very rare. In fact, there are no known absolute prime that contains 3 of them.
That is a class of toy problem because clearly it's not a property of the number itself but of the representation. For example, 73 is not an absolute prime in Binary but it's in base 8 but it's not in base 6. So we are really not learning anything about numbers but about dudes with 10 fingers. What is useful are the techniques for the proofs.
There are not so many absolute primes in binary representation, aren't there?
You can though apply the proof techniques in the video to any base to explore similar problems in those bases. So the proof techniques you’re learning here isn’t solely useful in base 10, modifications of it can be applied for other number systems.
@@Bodyknock Yes, e.g. in binary they may not have any zero, that means, you have a Mersenne prime, and all permutations are the numbers itself!
This is a point that, in my opinion, has many answers.
It is a bit like painting a still life. What value does it have when it has been done so many times before and to such a high standard of acclamation if we believe the acclaimers.
In other words it is not the result that matters. It is the doing of math.
There was a math researcher supposed to have something like 10 head and 21 legs. I forget the real numbers.
Author had a French name, based in France doing, I think, cutting edge math research