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Fonze

What is the next number set in the sequence?

What is the next number set in the sequence?  

33 members have voted

  1. 1. What is the next number set in the sequence? 012 023 034 045 056 067 078 089 ___

    • 090
      9
    • 091
      4
    • 101
      2
    • 112
      0
    • 123
      1
    • Other, post below
      17


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This is just a stupid curiosity with no right or wrong answer, but here's a mock-up situation to establish a coherent thought:

 

You are in a group, or at work, and multiple people use the same pw to log on to a system. For security reasons the pw needs to be changed every x days and cannot repeat the last 10, but for simplicity's sake the password stays the same, save a few numbers which follow a set sequence, so people don't have to be constantly notified of what the new pw is and can figure it out themselves.

 

The system you made started with 012, then 023, 034, 045, 056, 067, 078, and 089, but now it is time for the next pw, however the sequence could go in multiple directions and nobody agrees on what it should be. What do you set it to? 

 

There is a poll, but given that this is opinion-based with a lack of right and wrong, please provide an explanation for your choice. Looking forward to the read later :)

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i would pick 123 because although it seems wrong in terms of interval it's the only number where each digit is different!

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Other: 102, because the interval is similar and the digits are still all different.

Alternatively: It switches and comes back down the line but the last two digits are reversed, so 098 (followed by 087, 076, 065...)

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In my opinion, by logic, it should be 090 since the last 2 digits follows a continuous and progressive sequence but...

What if the group decides to implement a rule that forbids to repeat a digit in any position of the number, like 090?...well, then the next logical number should be 091 if it keeps following the sequence.

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I wouldn't use that system in the first place. Here's an idea that would cycle through all possible 3-digit number sets, but only ever changing one number at a time. That would make it different from simply counting. The sequence would go as follows:

000 001 002 ... 009 019 018 017 ... 010 020 021 022 ... (if I did it right) 900 000 ...

 

This is an decimal extension of Gray Binary Code. I just came up with it on the spot, but somebody probably has done it before.

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Rather than answer your question, I'll try to explain why there are infinite solutions. This will be done by showing that the set of all possible solutions is infinitely large.

 

So we use the Lagrange Interpolation Polynomial. For this case and for simplicity, this means that given a sequence of integers, we can construct a polynomial with rational coefficients has the values of the sequence when it is evaluated at integer values. (Like if we have a polynomial p(x); p(1) will give you the first term of the sequence, p(2) the second and so on).

 

Now consider the set for your question: {(1, 12), (2, 23), (3, 34), (4, 45), (5, 56), (6, 67), (7, 78), (8, 89), (9, n)}. I can define to be any integer (heck, I can define it to be any number - integer, rational, real, complex, etc.) and I can find a polynomial that satisfies it thus also giving justification. But we have hecking infinitely many choices for n, so any and every numerical answer is valid.

 

tl;dr - I couldn't help myself, so I proved that there are infinitely many solutions and justifications to the problem. Also yeah, now the password can be literally anything it wants while still being a logical (and unimaginably obtuse) change. So everyone wins!

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Either 090 or 100 (yes 100) could both work for me. 090 works as logicial progression compared to the given numbers, but 100 works as mathematical progression (x=11n+1 works for this sequence).

Edited by Scotty

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3 hours ago, Empyre said:

I wouldn't use that system in the first place. Here's an idea that would cycle through all possible 3-digit number sets, but only ever changing one number at a time. That would make it different from simply counting. The sequence would go as follows:

000 001 002 ... 009 019 018 017 ... 010 020 021 022 ... (if I did it right) 900 000 ...

Tangential to this, I ended up with a number system where, if you condensed all the numbers together, you'd end up with the fewest characters used overall (every number overlaps every other number as much as possible, meaning that each number only has a space footprint of a single character). I was trying to figure out the shortest way to enter every three digit number into a lock which had no clear button. I did this manually for two digit numbers during a lecture whilst I was bored, then generalised (though my generalisations were possibly unfounded, and definitely lacking in any sort of proof). If I remember correctly, you end up with 10n+n-1 digits (where n is 3 in this case). Here's all two digit numbers encoded in the fewest number of characters possible, which should give you a feel for what patterns might emerge in this system.

0010203040  5060708091  1213141516  1718192232  4252627282
9334353637  3839445464  7484955657  5859667686  9778798899 0

For three digits, you'd just go 0xy where xy goes through every two-digit number in order, so 089 would be followed by 090, though it would start with 0001 to avoid needlessly using characters. Things only get really interesting when you start messing with the most significant number, as it becomes more difficult to figure out what numbers have already been represented (obviously baking up an algorithm makes that easier).

 

On-topic, I'd personally go for

3 hours ago, Angry Saint said:

100, because you increase each time of 11, and 89+11 = 100

 

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1 hour ago, SGS Man said:

Rather than answer your question, I'll try to explain why there are infinite solutions. This will be done by showing that the set of all possible solutions is infinitely large.

 

So we use the Lagrange Interpolation Polynomial. For this case and for simplicity, this means that given a sequence of integers, we can construct a polynomial with rational coefficients has the values of the sequence when it is evaluated at integer values. (Like if we have a polynomial p(x); p(1) will give you the first term of the sequence, p(2) the second and so on).

 

Now consider the set for your question: {(1, 12), (2, 23), (3, 34), (4, 45), (5, 56), (6, 67), (7, 78), (8, 89), (9, n)}. I can define to be any integer (heck, I can define it to be any number - integer, rational, real, complex, etc.) and I can find a polynomial that satisfies it thus also giving justification. But we have hecking infinitely many choices for n, so any and every numerical answer is valid.

 

tl;dr - I couldn't help myself, so I proved that there are infinitely many solutions and justifications to the problem. Also yeah, now the password can be literally anything it wants while still being a logical (and unimaginably obtuse) change. So everyone wins!

*hits blunt*

 

Sorry I had to get on your level here real quick. Honestly that made no sense to me; I never studied numerical analysis, but I'm tickled at your response. The "answer" can logically go in many directions, such as mathematically +11 to 100 or sequentially 8->9 and 9->0 to 090. But this all misses the point of this question, as you and I have both said:

5 hours ago, Fonze said:

no right or wrong answer,

which would infer that even something like memfis's 666 could be subjectively the most logical conclusion to somebody. However none of this is what I'm interested in, i want to hear people's logic behind their personal choice; it's not the answer but how you get there. 

 

That said, I kinda would like to see some examples of random sets of 3 numbers being mathematically interpolated to logically fit the given set.

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@Fonze My point was that any answer can be mathematically justified by simply choosing a ninth value you want and providing a specific polynomial and then saying "This polynomial generates the first 8 terms of the sequence and also the ninth!"

 

For instance, this polynomial generates the sequence for the first 8 terms (that you gave) and predicts 666 as the ninth value: Ninth = 666

And this polynomial does the same but predicts 100 as the ninth value: Ninth = 100

And this one generates 123 as the ninth value: Ninth = 123

And this one generates 090 as the ninth value: Ninth = 90

And so on: Template (Replace n)

 

Simply put even Memfis' joke answer of 666 has a logical justification behind it. We simply construct a polynomial that generates your values in order and then 666. And we always have such a polynomial. For any ninth value you want.

 

So yeah, every answer is logical, even joke ones.

 

Quote

 i want to hear people's logic behind their personal choice; it's not the answer but how you get there. 

 

My answer would be 420 then because this polynomial justifies it: Ninth = Blunt

Edited by SGS Man

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15 minutes ago, SGS Man said:

@Fonze My point was that any answer can be mathematically justified by simply choosing a ninth value you want and providing a specific polynomial and then saying "This polynomial generates the first 8 terms of the sequence and also the ninth!"

 

For instance, this polynomial generates the sequence for the first 8 terms (that you gave) and predicts 666 as the ninth value: Ninth = 666

And this polynomial does the same but predicts 100 as the ninth value: Ninth = 100

And this one generates 123 as the ninth value: Ninth = 123

And this one generates 090 as the ninth value: Ninth = 90

And so on: Template (Replace n)

 

Simply put even Memfis' joke answer of 666 has a logical justification behind it. We simply construct a polynomial that generates your values in order and then 666. And we always have such a polynomial. For any ninth value you want.

 

So yeah, every answer is logical, even joke ones.

 

 

My answer would be 420 then because this polynomial justifies it: Ninth = Blunt

You have to consider that after the ninth value, people will need the n+1 value, still to be an integer number of 3 cifres to be used as code.

So actually you cannot have a whatever solution, you have an additional condition to satisfy and few of your proposal solutions satisfy it.

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12 minutes ago, Angry Saint said:

You have to consider that after the ninth value, people will need the n+1 value, still to be an integer number of 3 cifres to be used as code.

So actually you cannot have a whatever solution, you have an additional condition to satisfy and few of your proposal solutions satisfy it.

In such a case, yes there are a few valid answers and my answer would be incorrect in that interpretation. However, the question merely asks for the next term in the sequence and I have demonstrated that any three digit term is valid.

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If you dumdums used the Sixty Ninth Theorem of Calculus you'd realize the answer is 420

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If we want a 9th and 10th code, as Fonze wrote that the system remember the last 10 numbers, then it remains valid my above consideration.

 

If we want only a 9th code, I still belive it has to be something easy to remember, I don't mean as number but as way of generating it. I don't consider a 8th grade polynome with 9 terms to remember as easy, but I admit YMMV.

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30 minutes ago, Angry Saint said:

You have to consider that after the ninth value, people will need the n+1 value, still to be an integer number of 3 cifres to be used as code.

So actually you cannot have a whatever solution, you have an additional condition to satisfy and few of your proposal solutions satisfy it.

In such a case, yes there are a few valid answers and my answer could potentially be incorrect in that interpretation. However, the question merely asks for the next term in the sequence and I have demonstrated that any three digit term is valid.

 

Furthermore, my answer can be generalized easily, since there are several sequences that satisfy all the conditions and they are all mathematically justified. We only need to cycle 11 terms to satisfy all conditions.

 

Quote

If we want a 9th and 10th code, as Fonze wrote that the system remember the last 10 numbers, than it remains valid my above consideration.

 

If we want only a 9th code, I still belive it has to be something easy to remember, I don't mean as number but as way of generating it. I don't consider a 8th grade polynome with 9 terms to remember as easy, but I admit YMMV.

10

 

A simple case would be a periodic repetition of the {012, 023, 034, 045, 056, 067, 078, 089, 420, 666, 314} . No one needs to remember a massive polynomial but a completely random-looking sequence (that fulfills all the required criteria) is always logically justifiable. It may not be intuitive, but it will be logical.

 

 

Quote

If you dumdums used the Sixty Ninth Theorem of Calculus you'd realize the answer is 420

But the 420th Theorem of Algebra says it's 69 smh.

Edited by SGS Man : Okay wtf happened to my posting?

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10 minutes ago, 42PercentHealth said:

89+11=100

 

EZPZ

Yep, 100 is my answer. But, as pointed out, there are many correct answers. For the password issue, I might do something like this:

Spoiler

abc123

ab1c23

a1bc23

1abc23

ab12c3

a12bc3

12abc3

ab123c

a123bc

123abc

 

But wait, there's more!

ab123c

a123bc

a1b23c

1ab23c

1a23bc

a12b3c

 

etc.

 

With all of those, the letters, numbers, the order of the letters, and the order of the numbers stays the same - they are just intertwined enough to fool most "too similar" password checkers. And, I did not include all sequences - there's more of them. But that's not easy to remember.

 

If you can, use whole sentences:

"Here's 12 dollar signs: $$$$$$$$$$$$."

"I was born on August 14th!."

"An*asterisk*is*between*each*word"

 

Each of those are incredibly strong passwords (because of length), and include symbols, upper/lower, and the first two have numbers. And, they are easy to remember. Those garbage auto-generated passwords ($A23-x6@%b*c) are awful, and, because they are difficult to remember, someone will always write them down, making them insecure. Whole sentences are way too long to be brute-forced, so they are secure, and don't need to be written down. Only problem is that many systems stupidly put an arbitrary limit to password length. Aaargh, you can't win.

 

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Well......looks like this a proof that there are a LOT of answers with a LOT of changes to proceed when it comes for a password change protocol. 

Still, a fun way to exercise our brain with a sequence game...

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I had a dumb idea, could it be 000? Ignoring the non-repeating numbers, we could take into account that all the numbers begin with zero and could continue to begin with zero. 

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x(n+1) = x(n) + 11

Therefore, the next item in the sequence is 100.

 

It's the most "mathematical" solution, and I'm surprised it's not an option in the poll.

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9 minutes ago, Spie812 said:

x(n+1) = x(n) + 11

Therefore, the next item in the sequence is 100.

 

It's the most "mathematical" solution, and I'm surprised it's not an option in the poll.

It's also the most obvious solution, which is why it's presented as a trick question. Of course 89+11 is 100, but the last number is 089. In this scenario, the zero matters.

 

You're not wrong, but there seems to be more than one answer. It just depends on how you read the question.

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Well, it was the most obvious solution.

My favorite solution is to have 9s wrap around to 0. That way, the final digits go 78, 89, 90, 01, 12, and the sequence repeats itself.

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09A, then 0AB. Treating them as hex keeps you from changing the first digit, which the people in this office seem to have a problem with. Here you can continue to 0BC, start a new sequence at something like 112, or it's long enough now to where you can start over at 012.

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29 minutes ago, Spie812 said:

Well, it was the most obvious solution.

My favorite solution is to have 9s wrap around to 0. That way, the final digits go 78, 89, 90, 01, 12, and the sequence repeats itself.

Yes, there are many possibilities, including using something other than numbers (though the requirements state that the last 10 passwords must be unique). You could use hex, and the next number would be 09A.

 

For a 3-digit weekly password, you could treat year/month/day as a number (20180220), divide by 7, chop off decimals (~2882888), multiply by 11 (31711768), and grab the last 3 digits (768), and have a sequence very similar to the original question.

 

I have a similar issue at work, where I'm required to use upper/lower, numbers, and symbols, with a max length of 16. This thread has me thinking, as I too need to jot it down to remember it :(

 

EDIT: Heh, GuyMcBrofist ninja-posted about hex while I was typing this - nice!

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19 hours ago, kb1 said:

use whole sentences [as passwords]

Sir, you are a goddamn genius. Genius, I tell you.

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