What's special about 277777788888899? - Numberphile
video description
11-step record (277777788888899) first goes to 4996238671872, which obviously has 10 steps.
If you put that in ascending order (getting rid of the 1, you get 223466778899.
Obviously you could multiply the 2 2's to get 4, but then you can't multiply anything else, so it's inefficient. Actually, 9s, 8s, and 7s are the most efficient, so let's break down anything less than 7.
Break the 4 into 22 and the 6's in 23's, and you get 223(22(23(23)778899 (which is 222222333778899 after reordering)
More efficiently, group the 3's into 9's as much as you can, and the 2's into 8's as much as you can, and you get 3778888999. Which is the 10-step record.
We should really go the other direction if we want to find a new record breaker: find a number (let's start with any number, it doesn't matter for now if it's the lowest such number) whose digits multiply together to be 277777788888899.
SO all you have to do is factor 277777788888899.
Oh, it's prime? Okay, maybe rearrange the digits. You can at least find a number divisible by 32 (997, 777, 778, 888, 288, and you might have to rearrange the digits of the prime remaining then, but. it seems like this should work.
Or. couldn't you, any time you get a prime, add a 1 or two 1's to the beginning to make it divisible by 3?
Date: 2022-04-09
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Comments and reviews: 9
MLMI
Dear Matt & Nphile Friends - I call finding the -digital root- of a whole integer -summary digital reduction- (SDR, though I like calling the sum its -digital identifier- (Di, for the sake of disambiguation (re: other kinds of roots (of unity, infinity, etc. So, my questions are:
>
Q1. a. Should the result of checking for the numeric complexity of an integer's structural nature with the multiplicative digital reduction (MDR) process really be called its -multiplicative persistence- or its intrinsic (or internal) digital multiplicity? Q1. b. What (and how) might we discover what MDR could show us about Graham numbers x 10 x n-n?
>
Q2. Do y'all think that SDR and MDR could show us anything new and useful about
a. primals p
b. -(p1, 6n, p2)-
c. colossal 6n
d. other hyper-composite nonprimals (n, and
e. the relativity of primals/primality and dyadics/duality, etc.
?
>
Q3. Can MDR show us anything new and useful about the nature of numbers and/or maths?
> Thanks for another great episode on numbers & numeric logic.
reply
Dear Matt & Nphile Friends - I call finding the -digital root- of a whole integer -summary digital reduction- (SDR, though I like calling the sum its -digital identifier- (Di, for the sake of disambiguation (re: other kinds of roots (of unity, infinity, etc. So, my questions are:
>
Q1. a. Should the result of checking for the numeric complexity of an integer's structural nature with the multiplicative digital reduction (MDR) process really be called its -multiplicative persistence- or its intrinsic (or internal) digital multiplicity? Q1. b. What (and how) might we discover what MDR could show us about Graham numbers x 10 x n-n?
>
Q2. Do y'all think that SDR and MDR could show us anything new and useful about
a. primals p
b. -(p1, 6n, p2)-
c. colossal 6n
d. other hyper-composite nonprimals (n, and
e. the relativity of primals/primality and dyadics/duality, etc.
?
>
Q3. Can MDR show us anything new and useful about the nature of numbers and/or maths?
> Thanks for another great episode on numbers & numeric logic.
reply
Shihai-sha
Well I have an idea to get the highest Persistence world record with lowest digits
First of all start analyzing 10-1000 combinations of numbers then put in the best of the combinations as a higher random chance in you so explained program then give it an order to make the numbers in ascending order then just work on programming higher digit combinations to maximize all of it and if you want this part to be automatic too then leave it generating 5 digit combinations for atleast 60000 retries then go on to 7 digits and continue on
It'd really time consuming but its worth it cause you have to spend only 1 week at most to code it all and there you go world record is yours
reply
Well I have an idea to get the highest Persistence world record with lowest digits
First of all start analyzing 10-1000 combinations of numbers then put in the best of the combinations as a higher random chance in you so explained program then give it an order to make the numbers in ascending order then just work on programming higher digit combinations to maximize all of it and if you want this part to be automatic too then leave it generating 5 digit combinations for atleast 60000 retries then go on to 7 digits and continue on
It'd really time consuming but its worth it cause you have to spend only 1 week at most to code it all and there you go world record is yours
reply
AndreasFriedrich
It was quick to calculate. Time to extend the algorithm for more digits.
Tue Apr 5 10: 59: 34 CEST 2022: 3 -> 39-
Tue Apr 5 10: 59: 34 CEST 2022: 4 -> 77-
Tue Apr 5 10: 59: 34 CEST 2022: 5 -> 679-
Tue Apr 5 10: 59: 34 CEST 2022: 6 -> 6788-
Tue Apr 5 10: 59: 34 CEST 2022: 7 -> 68889-
Tue Apr 5 10: 59: 34 CEST 2022: 8 -> 2677889-
Tue Apr 5 10: 59: 34 CEST 2022: 9 -> 26888999-
Tue Apr 5 10: 59: 35 CEST 2022: 10 -> 3778888999-
Tue Apr 5 10: 59: 37 CEST 2022: 11 -> 277777788888899
reply
It was quick to calculate. Time to extend the algorithm for more digits.
Tue Apr 5 10: 59: 34 CEST 2022: 3 -> 39-
Tue Apr 5 10: 59: 34 CEST 2022: 4 -> 77-
Tue Apr 5 10: 59: 34 CEST 2022: 5 -> 679-
Tue Apr 5 10: 59: 34 CEST 2022: 6 -> 6788-
Tue Apr 5 10: 59: 34 CEST 2022: 7 -> 68889-
Tue Apr 5 10: 59: 34 CEST 2022: 8 -> 2677889-
Tue Apr 5 10: 59: 34 CEST 2022: 9 -> 26888999-
Tue Apr 5 10: 59: 35 CEST 2022: 10 -> 3778888999-
Tue Apr 5 10: 59: 37 CEST 2022: 11 -> 277777788888899
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Justin
Anyone else notice he wrote down two hundred and seventy-seven trillion, seven-hundred and seventy-seven billion, seven-hundred and EIGHTY-eight million, eight-hundred and eighty-eight thousand, eight-hundred and ninety-nine, but he said two hundred and seventy-seven trillion, seven-hundred and seventy-seven billion, seven-hundred and SEVENTY-eight million, eight-hundred and eighty-eight thousand, eight-hundred and ninety-nine at 0: 13?
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Anyone else notice he wrote down two hundred and seventy-seven trillion, seven-hundred and seventy-seven billion, seven-hundred and EIGHTY-eight million, eight-hundred and eighty-eight thousand, eight-hundred and ninety-nine, but he said two hundred and seventy-seven trillion, seven-hundred and seventy-seven billion, seven-hundred and SEVENTY-eight million, eight-hundred and eighty-eight thousand, eight-hundred and ninety-nine at 0: 13?
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Bandana_girl
I find it interesting that the optimal persistences for 6-11join the same track after the first step and are the optimized versions of each of the steps in the persistence route. With that, I doubt that the conjecture will hold that you would never beat 11 because surely, there must be some number in the infinite family of possibilities for the current 11 with prime factorization of the form 2-a-3-b-7-c
reply
I find it interesting that the optimal persistences for 6-11join the same track after the first step and are the optimized versions of each of the steps in the persistence route. With that, I doubt that the conjecture will hold that you would never beat 11 because surely, there must be some number in the infinite family of possibilities for the current 11 with prime factorization of the form 2-a-3-b-7-c
reply
Catman
tbh wouldn't it potentially become possible if we could find a number whose digits combine to 277777788888899 or any number with it's digits, or some other numbers that make it 11 steps?
i mean, i know it's gonna be an absolutely massive number to make it to this kind of value but it is theoretically possible?
it makes sense that there would be infinite numbers that make it 11 steps or in theory more
reply
tbh wouldn't it potentially become possible if we could find a number whose digits combine to 277777788888899 or any number with it's digits, or some other numbers that make it 11 steps?
i mean, i know it's gonna be an absolutely massive number to make it to this kind of value but it is theoretically possible?
it makes sense that there would be infinite numbers that make it 11 steps or in theory more
reply
nibblesdotbas
For anyone hard of hearing, at 3: 31 it should be -No, knock yourself out, as many many digits as you want. And often in these videos. - (current caption -No-Nock is about managers everyone and often in these videos-) Also, soon thereafter at 3: 42, part of the caption should be -as you may have noticed a little underprepared today- (not -as you may have those little underprepared today-.
reply
For anyone hard of hearing, at 3: 31 it should be -No, knock yourself out, as many many digits as you want. And often in these videos. - (current caption -No-Nock is about managers everyone and often in these videos-) Also, soon thereafter at 3: 42, part of the caption should be -as you may have noticed a little underprepared today- (not -as you may have those little underprepared today-.
reply
Mr
Why use recursion? You know how many digits there are, you've got each digit as an integer, and you've initialised your current product. Why not this?
product = 1
for each character
convert to digit
product = product - digit
next
(Please imagine whatever punctuation, layout, and spelling you like - if mine doesn't suit)
reply
Why use recursion? You know how many digits there are, you've got each digit as an integer, and you've initialised your current product. Why not this?
product = 1
for each character
convert to digit
product = product - digit
next
(Please imagine whatever punctuation, layout, and spelling you like - if mine doesn't suit)
reply
Arvid
You should be able to find a number that when you multiply them togheter you get that 277777788888899 or any number with that combination or any other number that have 11 steps and boom. You have 12 steps. So in other words. Prime factors of all the 11 steps number and then find one of them that have only 2, 3, 7, 11 as factors.
reply
You should be able to find a number that when you multiply them togheter you get that 277777788888899 or any number with that combination or any other number that have 11 steps and boom. You have 12 steps. So in other words. Prime factors of all the 11 steps number and then find one of them that have only 2, 3, 7, 11 as factors.
reply
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