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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Hi all

I have some simple but challenging enough problems to submit.

___Part I___

You are given a sequence of n digits and allowed to split it by grouping at most 2 consecutive digits together.

Example: 796421 may give the splitting 7 96 4 21

***Find a formula for the number of possible splittings.***

___Part II___

You are given a number X and a sequence of digits.

Your task is to split the sequence so that the obtained groups

(of at most 2 digits) add up (or average to) X.

Examples

81 as a sum for 15723 1 57 23 --> 81

27 as a mean for 15723 1 57 23 --> 81/3 = 27

Can you solve the following:

77 as a sum for 15683

79 as a sum for 532168.

88 as a sum for 191414941

31 as a mean for 9579

14 as a mean for 884653

42 as a mean for 11398917

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hi meanmaths

Welcome to the forum!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

Part 2)

77 = 1 + 5 + 68 + 3

79 = 5 + 3 + 2 + 1 + 68

88 = 19 + 14 + 1 + 4 + 9 + 41

31 = ( 9 + 5 + 79 ) / 3

14 = (8 + 8 + 46 + 5 + 3) / 5

42 = (11 + 3 + 98 + 91 + 7 ) / 5

Part 1 is more interesting.

By playing STP

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hi bobbym

What is n=F[n+1] supposed to be? As far as I know, the only solutions to that are 1 and 5.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I meant

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Congrats anonimnystefy and bobbym

Your quick answers are impressive, considering that:

- I posted Part II for days in the Mathematics community (~70k) of Google+ and only the first 2 got solved.

- I also posted Part I yesterday before posting it here. still waiting for an answer.

Let's continue this.

__Part III__ (tougher cookie than Part I but I have even tougher ones in case someone manages to solve this.)

You are given a sequence of n digits and allowed to split it by grouping at most 2 consecutive digits together.

Example: 796421 may give the splitting 7 96 4 21

Difference with Part I is that

An orientation is attached to 2-digit splits, meaning they can be read either from left to right or right to left

So 7 96 4 21 (in the above example) may correspond to 7 96 4 21 or 7 96 4 12 or 7 69 4 21 or 7 69 4 12

Whether the two different orientations give the same number or not is not taken into account:

a split 33 (left to right) is different from a split 33 (right to left) because the pairs [2 consecutive number][orientation]

are different

***Find a formula for the number of possible splittings.***

__Part IV__

You are given a number X and a sequence of digits.

Your task is to split the sequence so that the obtained groups

(of at most 2 digits) add up (or average to) X.

You can define your splits from left to right

Examples

24 as a sum for 3381 3 3 81[right to left] --> 3 3 18 --> 24

15 as a mean for 9423 9 4 23[right to left] --> 9 4 32 --> 45/3 = 15

Solve the following:

177 as a sum for 3534477

204 as a sum for 944949165

34 as a mean for 691795

35 as a mean for 59458531

Constraint: Use at least one backward orientation.

In this series, parts with odd numbers will be about formulas

while parts with even numbers will propose some (more and more) challenging

puzzles based on number splitting.

[To the moderator, let me try to convince you not to cut the following :-)

The challenges are taken from the game, which is a free (without any in-app purchases) maths game that may interest people who try to

solve the challenges. I'm a (hobby) app developer but I'm also a computer science Ph.D. with a good day-job. So this is not about money.]

More (diverse and randomly generated) puzzles await you in the Android game

MEAN SUMurai. The game includes an online versus mode, achievements and leaderboards.

If you take a look, you will know that very serious math challenges await you

in the next parts.

Best

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

let me try to convince you not to cut the following

I will take you on you word that it is not for monetary profit. In return I ask that you place a link to this site on your page.

Part IV)

177 = 53 + 43 + 74 + 7

204 = 49 + 49 + 49 + 1 + 56

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Hi bobbym

Congrats for the two sums.

I just placed a link to the forum on the game's official website.

meansumz com

All this is a part-time one-man effort, so, as you can see, no fancy stuff.

You look pretty good at number splitting. I have some insane but solvable instances in the game

that you may like. Take a look at the last picture in the download page of

Mean Sumurai in the Google Play Store.

This is one of the toughest instances I solved, in exactly 59 s.

[I integrated in the game the ability to save solved instances.]

Though I'm reasonably good at Maths, I'm far from being a "human calculator".

The secret: a strong mental heuristic that I developed through many many hours of

playing.

Note that the app does contain ads but it mainly allows me to get an idea of how much the game is played.

Right now, the game is played in 22 countries and though there are less than 100 players (I launched the game

4 days ago), I take comfort in the fact that there is always at least one person playing it :-)

Best

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I have some questions on part III.

Whether the two different orientations give the same number or not is not taken into account:

a split 33 (left to right) is different from a split 33 (right to left) because the pairs [2 consecutive number][orientation]

are different

I am not following that. Are you saying you want each pair counted twice?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

bobbym wrote:

...Are you saying you want each pair counted twice?

You can see it that way, as long as you don't forget that you're looking for the number of combinations.

So, as you probably already know, if you have n pairs in a splitting, it does not mean that it counts for 2*n.

Edit: Just in case: "pair" in what I wrote above refers to 2-digit groups

*Last edited by meanmaths (2013-08-02 04:25:31)*

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

In the case of pairs that were of the type (2,2),(3,3), (1,1) though would always be counted twice?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

yes. Otherwise, a formula would be much tougher to find.

You can see it the following way. Each digit is assigned a number representing its position in the sequence.

So regardless of the actual digit, a position is always unique. So, if you have 888 as a sequence,

a splitting 8 88[left to right] would correspond to {position1, (position2, position3)}

while a splitting 8 88[right to left] would correspond to {position1, (position3, position2)}

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hm, is 21 1 2 a valid splitting of 1221?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

anonimnystefy wrote:

Hm, is 21 1 2 a valid splitting of 1221?

hmm 21(12 backwards) 2 1 is a valid splitting and, because the splits do not have to be ordered, so is 21 1 2.

In an attempt of making things more clear, let me propose an alternative formulation of the problem.

The sequence is like a series of dots, and each dot can either be alone or or connected to

its predecessor or successor.

so let's take 1221

(dot1) (dot2) (dot3) (dot4)

21 2 1 corresponds to

(dot1)<-(dot2) (dot3) (dot4)

and because there is no order for the splits

your 21 1 2

(dot1)<-(dot2) (dot4) (dot3)

is also valid, though possibly confusing.

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hm, so, we count 21 1 2 and 12 1 2 as one splitting, but 12 1 2 and 12 2 1 as two, right? And is 3 12 4 a valid splitting if 1234, then?

Here lies the reader who will never open this book. He is forever dead.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

anonimnystefy wrote:

Hm, so, we count 21 1 2 and 12 1 2 as one splitting, but 12 1 2 and 12 2 1 as two, right? And is 3 12 4 a valid splitting if 1234, then?

No, you can't count 21 1 2 and 12 1 2 as one splitting. 2->1 is definitely different from 1->2.

hehe, this is a perfect case-study of why anticipating questions is not always a good thing.

I just confused you guys. Forget totally about same digits stuff and just focus on the following

Consider that you are given a sequence of 4 digits

Assign these digits numbers from 1 to 4.

Below, an enumeration of the all different combinations.

1234

1 2 3 4

1 2 34 -- 1 2 43

1 23 4 -- 1 32 4

12 3 4 -- 21 3 4

12 23 -- 12 32 -- 21 23 -- 21 32

so 11 different splittings for n = 4

*You don't need to care about whether digit 1 is equal or not to digit 2.

*Order is irrelevant. While looking for a formula, you may write 1 2 4 3 instead of

1 2 3 4. It is the same splitting but I highly suggest you stick with the order in

which the digits appear. Otherwise, you may get unnecessarily confused.

[If you want to introduce order, you can do so and it could be interesting

but it is outside the scope of the challenge]

Now, if you really want to see the context of all this, go download the game

Mean Sumurai @ Google Play. This could sound like another shameless plug.

It may well be one :-) but I sincerely think it could help you guys.

Also, don't think this is an easy problem. It is pretty hard and though it has a clean answer

it is not something like Fn+1. It is slightly messier.

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hi meanmaths

*Last edited by anonimnystefy (2013-08-02 23:14:52)*

Here lies the reader who will never open this book. He is forever dead.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Congrats anonimnystefy

You're really skilled :-). I'll let some time (may be a day or 2) for others to possibly come with an answer.

Then I will continue the series with some even harder challenges (formulas + games) on the same theme.

Meanwhile, you can get hints about what's coming by taking a look at MEAN SUMurai (@ Google Play).

Also, don't forget this part of the challenge

"

34 as a mean for 691795

35 as a mean for 59458531

with a least one split to read from right to left.

"

Yeah, I know it's less cool than finding formulas but it requires different skills and qualities.

Best

PS: Latest, more clear version of the challenge below for people who want to try

[CHALLENGE]

Consider that you are given a sequence of n=4 digits

Assign to these digits numbers from 1 to 4.

Below, an enumeration of the all different combinations.

1234

1 2 3 4

1 2 34 -- 1 2 43

1 23 4 -- 1 32 4

12 3 4 -- 21 3 4

12 23 -- 12 32 -- 21 23 -- 21 32

so 11 different splittings for n = 4

Find a formula for the number of combinations, given n.

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Thanks! I think I'll leave part II for someone better than me at spotting stuff.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Hi bobbym

You don't have it for now.

Your formula must allow computation of

the number of combinations for any n.

Cheers

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

I worked on this one

Also, don't forget this part of the challenge

34 as a mean for 691795

35 as a mean for 59458531with a least one split to read from right to left.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

oh sorry. probably a lack of sleep.

Great Job bobbym! I'm definitely impressed by this forum. You're serious about your fun :-)

Well, I'll get some rest, go out a bit and probably post Parts V and VI

by this evening or tomorrow morning.

Best,

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Have a good night.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Thanks bobbym

It was actually the afternoon and I had only 2-3 hours of sleep the night before.

I love my little boy more than anything but he is not exactly helping with my on-off insomnia problems :-).

Back to business

New people on the thread, please go read the first posts. Otherwise, you may not get what's going on.

We've been discussing a new(?) kind of splitting and I've been trying shameless plugs of Mean Sumurai,

an android game based on it ;-D. bobbym and anonimnystefy have solved all my challenges up to now.

___________Part V: The formulas

Given a sequence S of digits, and splits of at most 2, what is the number of possible

combinations if we impose

(a) 1 digit to be left out of the splitting

(b) 2 digits to be left out of the splitting

Problem is that the position of the noise digits affects the number of possible splits

because a split is always made of consecutive digits.

Let me give you examples with a sequence of length 6

We have as positions 123456

You have to consider that, every position, if left out, defines a different space of combinations.

For instance, if left out, the digit at position 1 defines the sequence 23456 (5 digits)

which, as we saw in Part I, gives F6 combinations.

Whereas the digit at position 3 would define 2 distinct sequences 12 and 456

and thus F3 * F4 combinations

Good Luck! You will need it ;-)

[F for Fibonacci]

________Part VI: The games

Example:

120 as a sum for 834978 --> 8+34+78 = 120 (9 left out)

Challenges

110 as a sum for 872992 (Leave out 1 digit)

92 as a sum for 43971296 (Leave out 2 digits)

17 as a mean for 771133843 (Leave out 1 digit)

23 as a mean for 497353 (Leave out 2 digits)

52 as a mean for 5989652 (Leave out 2 digits)

In Mean Sumurai, the games are presented in a more gamer-friendly way,

along with (depending on the difficulty level) a calculator option that

displays the current computations. Text-version like the ones above are probably harder

to solve but they are doable.

Best,

****************************************************************************

[Note for bobbym and anonimnystefy: there are no right to left considerations in these parts]

****************************************************************************

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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