[Guest]

In a far² parallel dimension... on a tiny pink planet, Zearth, evolve the first 2 intelligent bacterial beings, Fred and Frank. In fact they are so intelligent, they even know that they are the only life that exists. Each "intelligent bacterium" splits into two at the end of each day. Life in this universe can not die.

On day 1, Fred creates the universe's first telephone. Fred also appoints himself the universal telephone operator and assigns himself the number "0". He assigns Frank the number "1". On day 2, two new members are introduced to the universe, Fred assigns the new members telephone numbers "2" and "3".

On Day 3, there are 8 beings.... On day 4, there are 16 beings, and Fred realizes a PROBLEM, bacteria can't tell up from down! So #6 and #9 would keep getting each others calls. So he decides to avoid both.

Summary of the facts.

1) Day one > 2 bacterium, day two > 4, day three > 8, day four 16..... so on

2) Bacterium can't tell up from down (Fred avoids all flip-able numbers.. '6' and '9' are out, '161' and '191' are out)

3) Every bacterium gets a telephone number (1,2,3,4,...1243.....123456...)

Question: What is the sum of all the telephone numbers in the universe on the 100th day.

Bonus: If you are a bacterium spawned on the 101th day, what is the chance your telephone number will have at least one 6 in it.

[Glycereine]

I'm equally confused which is why I figured 10 was ok... since there wouldn't be a 01.

I don't know where to go anymore because I don't think I even understand the numbering system now...

[Guest]

I'm equally confused which is why I figured 10 was ok... since there wouldn't be a 01.

I don't know where to go anymore because I don't think I even understand the numbering system now...

1 is different from 01.... Fred would assign 1, but will not assign 01 as if read upside down it would read 10, just to be safe he will avoid 10 too. Basically any number that can be reversed and have a different meaning. Since it is a telephone number we do not ignore those starting 0.

Edited June 9, 2009 by adiace

[Glycereine]

1 is different from 01.... Fred would assign 1, but will not assign 01 as if read upside down it would read 10, just to be safe he will avoid 10 too. Basically any number that can be reversed and have a different meaning. Since it is a telephone number we do not ignore those starting 0.

So can 010 be assigned in the place of 10? I'm basically trying to figure out which exclusions exist and I thought I had a pattern but I actually forgot about 0 as well as I eliminated numbers like 609 (which is ok to flip) before realizing they'd be ok.

edit: As soon as I posted that I realized 609 is probably not ok because unless these bacteria actually flip their numbers 180 degrees and not just upside down, 609 would turn into 906 .

edit2: Nevermind... as far as I can see 609 should be ok. Because with it upside down (not 180) it's not a number.

Edited June 9, 2009 by Glycereine

[k-man]

1 is different from 01.... Fred would assign 1, but will not assign 01 as if read upside down it would read 10, just to be safe he will avoid 10 too. Basically any number that can be reversed and have a different meaning. Since it is a telephone number we do not ignore those starting 0.

Ok, I'm sorry. I used a flippable number in my example. Let me restate my question.

How does Fred assign someone a number 02 or 002 or 000002? Basically the question is what is the sequence of assigning the numbers? Which number will be assigned before and after 02 for example?

Edited June 9, 2009 by k-man

[Glycereine]

Ok, I'm sorry. I used a flippable number in my example. Let me restate my question.

How does Fred assign someone a number 02 or 002 or 000002? Basically the question is what is the sequence of assigning the numbers? Which number will be assigned before and after 02 for example?

Well crap, I think after you posed your question I understood it a little better.

Let me take a stab at the numbering system.

0,1,2,3,4,5,7,8,00,02,03,04,05,07,11,12,13,14,15,17,20...

0 as a digit, it comes after nothing and before 1?

[k-man]

Well crap, I think after you posed your question I understood it a little better.

Let me take a stab at the numbering system.

0,1,2,3,4,5,7,8,00,02,03,04,05,07,11,12,13,14,15,17,20...

0 as a digit, it comes after nothing and before 1?

I was thinking about this too, but I wanted to hear it from adiace. This system also means that after we exhaust all 2-digit numbers (after 97) the next numbers will be 000, 002, 003, etc, so the whole thing starts all over.

[Glycereine]

I was thinking about this too, but I wanted to hear it from adiace. This system also means that after we exhaust all 2-digit numbers (after 97) the next numbers will be 000, 002, 003, etc, so the whole thing starts all over.

Yup. I'm working it that way now to see if there's a better pattern than there was without the 0's (it was terrible without).

Hopefully Adiace gets back to us soon to confirm or deny .

[Guest]

Yup. I'm working it that way now to see if there's a better pattern than there was without the 0's (it was terrible without).

Hopefully Adiace gets back to us soon to confirm or deny .

I think you have understood it right.

n = 1: 10 possible numbers (0 to 9), now remove the trouble one 6 and 9 > 8 possible telephone numbers

n = 2: 10*10 = 100 possible numbers (00.. 01.. 02.. 99), remove the trouble numbers (01, 06, 66, 80... etc..) but keeping 00, 11 and 88 > 100 - 25 + 3 = 78

so if you go up to 2 digits we can have 86 telephone numbers

[Guest]

well the synergistic (im just making up words but spell check actually liked it) numbers would be

5^([n/2]) []ceiling i think for each level but i might be a lil off my logic is choose half the numbers the others are automatic copies of the front so you choose the front half and the rest of the number is done. I havent done any of the problem tho just happened to think of that much so maybe 5^n-5^[n/2] for each n digits but i dont know how much this helps havent thought that far

[Glycereine]

I think you have understood it right.

n = 1: 10 possible numbers (0 to 9), now remove the trouble one 6 and 9 > 8 possible telephone numbers

n = 2: 10*10 = 100 possible numbers (00.. 01.. 02.. 99), remove the trouble numbers (01, 06, 66, 80... etc..) but keeping 00, 11 and 88 > 100 - 25 + 3 = 78

so if you go up to 2 digits we can have 86 telephone numbers

Just so I don't do even more math incorrectly, I have 80 numbers for 2 digits. I'm assuming the two I kept that you didn't were 69 and 96. Flipped they are both still the same.

[Guest]

Just so I don't do even more math incorrectly, I have 80 numbers for 2 digits. I'm assuming the two I kept that you didn't were 69 and 96. Flipped they are both still the same.

Good catch your right.

[Guest]

well the synergistic (im just making up words but spell check actually liked it) numbers would be

5^([n/2]) []ceiling i think for each level but i might be a lil off my logic is choose half the numbers the others are automatic copies of the front so you choose the front half and the rest of the number is done. I havent done any of the problem tho just happened to think of that much so maybe 5^n-5^[n/2] for each n digits but i dont know how much this helps havent thought that far

If i understand you.. 5^(n/2) are symmetric numbers... (symmetric numbers are.. 11, 888, 619, 6618199 and the like..)

so if n =2 .. then there are 5^1 = 5

if n = 3 then 5^1.5 ?.... what do you mean by ceiling? 5^2 ? then there should be 25 symmetric 3 digit numbers?

[Glycereine]

For n-digit phone numbers,

5^n numbers contain all 0,1,6,8 and 9's.

For Odd "n":

The symetric numbers are 3*5^((n-1)/2)

The excluded numbers are 5^n-3*5^((n-1)/2)

so for 1-digit numbers there are 5 that contain nothing but 0,1,6,8,9 but 3 of them are symetric, so the number excluded is 2. For 3 digit numbers there are 125 with nothing but 0,1,6,8,9 and 15 are symetrical, so the number excluded is 110.

For even "n", the symetric number of numbers is 5^(n/2).

So for 2 digits, there are 25 numbers with nothing but 0,1,6,8,9 and 5 are symetric, so 20 are excluded.

For 4 digits, there are 625 numbers with 0,1,6,8,9 and 25 are symetric, so 600 are excluded.

This holds true forever. Working on the actual problem now.

Edited June 9, 2009 by Glycereine

[Guest]

my answer was slightly wrong if there are an odd digit of numbers then the center digit cannot be 6 or 9 to be symetric because those numbers upside down arent themselves so the correct answer to how many numbers cant be used in n digit numbers is

5^n-((5^[n/2])*((3/5)^((n)mod2)) []ceiling or similarly 5^n-((5^[n/2])((3)^((n)mod2))[]floor

celing means round up no matter what [1.1]=2 [1.9]=2 [3]=3

floor just means round down

mod for people that dont know is remainder so nmod2 would be 1 if it is an odd number and 0 if it is even

so this in english is 5^n numbers with only 01698

5^[n/2]ceiling would be the subset of numbers that can be used because they are the same right side up and up side down but you need to eliminate 6 and 9 from being the middle number so multiply by 3/5 if odd digit because you used to have 5 in the center but now you can only have three

for exampl 1881 good 1691 good 16891 good but any number with --6-- or --9-- (-- being any quantity of numbers) cannot exist because the closest --6-- could be to being synergisitic is --9-- upside down. so it can be

anyway so 10^n would be total numbers

so

10^n-(5^n-((5^[n/2])*((3/5)^((n)mod2)) []ceiling would be usable numbers so for 2 digits

10^2-(5^2-((5^1)*(3/5)^0))=100-(25-5)=80

10^3-(5^3-((5^[2])*((3/5)^1))=

1000-(125-((25*3/5)))=890

im assuming you guys have found the 80 in the two digits to confirm that this works at least on that case and in my mind the logic works on all cases im pretty sure once this is accepted the summation of these numbers is quite simple but i havent really looked at that yet so it may turn out not to be

any questions,comments,wtf's

[Guest]

ah crap didnt see he ran with my idea before i did

[Glycereine]

my answer was slightly wrong if there are an odd digit of numbers then the center digit cannot be 6 or 9 to be symetric because those numbers upside down arent themselves so the correct answer to how many numbers cant be used in n digit numbers is

5^n-((5^[n/2])*((3/5)^((n)mod2)) []ceiling or similarly 5^n-((5^[n/2])((3)^((n)mod2))[]floor

celing means round up no matter what [1.1]=2 [1.9]=2 [3]=3

floor just means round down

mod for people that dont know is remainder so nmod2 would be 1 if it is an odd number and 0 if it is even

so this in english is 5^n numbers with only 01698

5^[n/2]ceiling would be the subset of numbers that can be used because they are the same right side up and up side down but you need to eliminate 6 and 9 from being the middle number so multiply by 3/5 if odd digit because you used to have 5 in the center but now you can only have three

for exampl 1881 good 1691 good 16891 good but any number with --6-- or --9-- (-- being any quantity of numbers) cannot exist because the closest --6-- could be to being synergisitic is --9-- upside down. so it can be

anyway so 10^n would be total numbers

so

10^n-(5^n-((5^[n/2])*((3/5)^((n)mod2)) []ceiling would be usable numbers so for 2 digits

10^2-(5^2-((5^1)*(3/5)^0))=100-(25-5)=80

10^3-(5^3-((5^[2])*((3/5)^1))=

1000-(125-((25*3/5)))=890

im assuming you guys have found the 80 in the two digits to confirm that this works at least on that case and in my mind the logic works on all cases im pretty sure once this is accepted the summation of these numbers is quite simple but i havent really looked at that yet so it may turn out not to be

any questions,comments,wtf's

My only comment is that 6 or 9 can be the middle number as long as the number still has 2,3,4,5 or 7 anywhere else in the number. I think I understand what you mean though, with managing symetric numbers, you multiplied the symetric number by 3/5 to eliminate 6 and 9 in the middle on odd numbers of digits. If so then you have the same thing I do. As for the other numbers, I worked it out for 80, 890 and 9400 to check my math then started using the equations.

[Guest]

yah the whole time i was only talking about the 10896 numbers anyway

i found a way to sum it but it prolly would need a computer but my logic might help someone else so for n digits the summation of all numbers is simple but for these numbers to take out it would look like this so far

this would be run in a loop for (whatever digit it goes to i didnt look if anyone bothered to find that) >n>0

this would be run in a loop of n=#of digits go untill n=0

(10^n*5^(n-1)*24+10^(n-1)*5^(n-1)[24]+10^(n-2)*5(n-1)(24)+....)-(10^n(5^((N/2)(floor))*(3^nmod2)

actually let me simplify this first

5^(n-1)24(10^n+10^(n-1)+10^(n-2)+...)-((5^([N/2](floor))*(3^nmod2))24(10^n+10^(n-1)+10^(n-2)...)

both done till n=0 then you need to remove the bad add you made for the non symetrical numbers with 6 and 9 in the center so same as mod if its odd

if((n/2)(ceiling)!=(n/2)(floor))

{

subtract 10^(n/2)[ceiling](5^(n/2)[floor])(3^nmod2)(15)

}

this part says if its an odd digit number then subtact that digits values added (24 of them) and add back the ones that arent 6,9 (24-15=9) im not sure this works ill work on it in between periods but i think it is promising

but it might not make much sense to anyone but me, so ill program it and show u guys data from it

[Glycereine]

That looks promising.

It goes to the 31st digit... However it only uses 14.24% of the 31 digit numbers and I'm not really sure how to account for that. Not sure if you can use 30.1424 as the number of digits .

This is what's killing me on the 6's also. The 6's used previously digits 0-29 are irrelevant, but some of the 30 digit numbers are used for generation 101 and of the ones that are used alot start with 9 (and therefore some end in 6). Not to mention all the pairs of 6/9 and 9/6 in the middle. Needless to say... I'm working on it.

[Guest]

yah i wrote a computer program to do that and it overloaded quickly and then had it find each individual digit seperately and each digit overloaded. and this is just what to subtract so I found a semi close substitute but no offense im starting to doubt there is a concrete solution. who started the thread was it adiace?

[Guest]

yah i wrote a computer program to do that and it overloaded quickly and then had it find each individual digit seperately and each digit overloaded. and this is just what to subtract so I found a semi close substitute but no offense im starting to doubt there is a concrete solution. who started the thread was it adiace?

There's a bit of a tricky part to doing it with a computer in that a 64bit long long isn't large enough to hold a 2^100 value. You might have some luck with a library that does character-based math for large numbers like a human would.

For those going after it manually, there's a decent looking online calculator for large numbers here: Calculator.

[Guest]

obviously but the point is even if your just considering the numbers to remove your talking about the value of the 30 digit numbers (someone, glycerine i think, said it ended around their im taken that for granted) anyway the value for each digit would be 5^29 *24 (well minus some stuff but still thats the beggest term so mid is prolly 5^29) but im working on other math just started working again

[Guest]

ok this is all ive got then im gonna need some input

this is only for if the numbers stopped at 30 digits and filled all of the 30 digits but this is just to show the scope of numbers we're eliminating

so each odd digit excluding 29 and 1 = you add this

-24(5^29)+24(5^15)+9(5)^14+24(5^(x-1))+24(5^((x-1)/2))-9(5)^(((x-1)/2)-1) x is the digit i.e. 1000 place is...3 or 4... 4 i think

anywho for even digits its a lil different but not much

-24(5^(29))+24(5^15)+9(5)^14+24(5^x-1)-33(5^((x/2)-1)

x is digit

now most of this math is based on the fact that say a digit of a 0to30 digit number

it is in 24(5^30+5^29+5^28+...+5^x) multiply this by 1-1/5 gives you 5^30-5^x-1 redivide

(5^30-5^x-1)/(4/5) and that is a main component now for remove synergies and so on

now i doubt this will really help but the point is that number times 10^28 number applied to x=28 times 25 should be close to what were talking about (reasoning more should be deleted after the times 25 but alot more should be added to so extremely estimating estimation)

and then that is just what you eliminate the rest is about 2^29 times that .. now this is all supposition but im done untill more info comes to light

[Glycereine]

Yeah I don't have it in me to continue down this path unfortunately. I worked on it for a good chunk of today and there's not a pattern that I can see that makes this something I can calculate without brute forcing it.

I mean we did establish patterns for how many numbers to eliminate for each set of digits, but that doesn't help immensely when the number of digits don't correspond evenly with a generation of the bacteria.

Oh well. On to other puzzles.

[Guest]

Yeah I don't have it in me to continue down this path unfortunately. I worked on it for a good chunk of today and there's not a pattern that I can see that makes this something I can calculate without brute forcing it.

I mean we did establish patterns for how many numbers to eliminate for each set of digits, but that doesn't help immensely when the number of digits don't correspond evenly with a generation of the bacteria.

Oh well. On to other puzzles.

You are right it is a lil off-putting, my spreadsheet program handles 50 digits so it was a breeze

Let me modify the question: to 20 days.

Also: If Fred is assigning 3 digit numbers... he will start with 000...002...010...012..097.. and end at 997.

1) What is the number allocated to the last bacterium on the 20th day.

Edited June 10, 2009 by adiace

[Glycereine]

You are right it is a lil off-putting, my spreadsheet program handles 50 digits so it was a breeze

Let me modify the question: to 20 days.

Also: If Fred is assigning 3 digit numbers... he will start with 000...002...010...012..097.. and end at 997.

1) What is the number allocated to the last bacterium on the 20th day.

The 1048576th bacteria (last spawned on day 200) will have the telephone number of 954247