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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.

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

hmmm..... the 1st and 2nd digits match my ans... could u support your ans?

Also: there is a way to achieve the ans to the original questions as well.

1) Sum of all telephone numbers on the 20th day

2) Chance your telephone number will have at least one 6 in it if you are born on the 21st day.

Edited by adiace
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hmmm..... the 1st and 2nd digits match my ans... could u support your ans?

Also: there is a way to achieve the ans to the original questions as well.

1) Sum of all telephone numbers on the 20th day

2) Chance your telephone number will have at least one 6 in it if you are born on the 21st day.

for the 20th generation of bacteria, there will be 1048576 bacteria (2^20).

When just 1 digit numbers are used, 8 phone numbers exist.

With 2 digit numbers included it raises to 88.

3 - 978

4 - 10378

5 - 107328

6 - 1091828

This is derived from previous discussion about exclusions.

So for the 1048576's bacteria, we will have used all but 1091828-1048576 available phone numbers up through 6 digits.

This is equal to a remaining 43252 phone numbers which mean's the 1048576th bacteria's phone number is 43252 numbers from the end of the 6 digit numbers.

Since this number is only 5 digits, we know that the first digit of the number is 9 which means there will be exclusions (flippables) in this section of phone numbers. Therefore we have to eliminate those that are flippable. We can determine that for each ten-thousand's place that begins with 0,1,6,8 or 9, there will be 625 flippable numbers, 5 of which are symetrical and can be included:

-- 5^5 for flippable numbers of 5 digits. For symetry, note that the 5 digit number is preceeded by 9 and must end with 6. So for each option of 9,8,6,1,0 in the 10,000 place, only 5 are symetrical. So there are 620 exclusions per 9,8,6,1,0 in the 10,000 place.

So we work backwards, knowing that we have 43252 remaining phone numbers, to work through but there are exclusions left out,

43252-9380 gets rid of the numbers that contain 99xxxx leaving us with 33872 numbers.

Then 33872-9380 gets rid of the numbers that contain 98xxxx leaving us with 24492 remaining numbers.

7 in the 10,000 digit means no flippable numbers, so 24492-10000 gets rid of the 97xxxx numbers, leaving us with 14492 remaining numbers

14492-9380 gets rid of the 96xxxx numbers, leaving us with 5112 numbers.

luckily 5 in the 10,000 digit also means no flippable numbers, so starting from 959999 we subtract the remaining unused numbers, 5112

and we have 954887 as the number for the 1048576th bacteria

note: this answer is different than my previous one but I made an error in my spreadsheet regarding symetry

I leave the other two questions to someone else.

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@ G

"So for the 1048576's bacteria, we will have used all but 1091828-1048576 available phone numbers up through 6 digits."

1091828>1048576 you have more than required numbers.

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@ G

"So for the 1048576's bacteria, we will have used all but 1091828-1048576 available phone numbers up through 6 digits."

1091828>1048576 you have more than required numbers.

Yes, there are more available numbers (up to 6 digits) than bacteria. Which is why I worked backwards from the highest number available (of 6 digits) based on the difference between numbers available and the number needed.

Result was in my above post along with reasoning along the way for exclusions or not.

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hey who said that it had to be 30 digits i just decided to look into that fact and it can be done in 10ish am i retarded or something or is the 30 digits thing wrong

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edit - ignore most of this good process by the way

i may be wrong but

-

" 5^5 for flippable numbers of 5 digits. For symetry, note that the 5 digit number is preceeded by 9 and must end with 6. So for each option of 9,8,6,1,0 in the 10,000 place, only 5 are symetrical. So there are 620 exclusions per 9,8,6,1,0 in the 10,000 place."

shouldnt it de 5^5-5^2=3100

because there are 5^5 numbers with only 01689 and then you add back in the symetrical ones which must start with 9 and end with 6 so you can choose two of the numbers of the remaining four and then the rest are done

what you found is the numbers that start and end with 9 and 6 respectively you then have to find out which one have symetrical insides too

..i think this is right

anyway i like how well we both moved on from this problem

p.s. i think i might be able to solve it pretty easily if it really only goes to 9 digits so please hurry and tell me if im wrong about that

Edited by final
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by similar logic can we say 2^30 last number would be 959836495

because you need 1073741924

in 9 digits you can get 1108672828

so the distance is 34931004

which there are 77500 eliminated in the 90000000

same in the 8000...

0 in 7000...

77500 in 600000

so you end up with 163505 left to go into the 959999999

ideas comments

plus ignor the crap i said about the 5^29 a while ago i was under the understanding that someone had decided that we were in the 30th digit and i didnt question it.. i really should of how could 2^30 be any where near 10^30... but tired minds do tired things so what u gonna do.

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ok so i reran my program for 9 digits (instead of 29 such an idiot) and got the sum of all the exclusions up to 10 digits or 999999999 to be 105775090316784 so i need to figure out how to subtract the ones at the end not excluded then we are done

if this is right

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Yes, there are more available numbers (up to 6 digits) than bacteria. Which is why I worked backwards from the highest number available (of 6 digits) based on the difference between numbers available and the number needed.

Result was in my above post along with reasoning along the way for exclusions or not.

From what I understood... you start with max possible in 6digits then fixed 9, so its 9XXXXX and subtracted your calculation for 5 digits.. but when you do that, you must keep in mind the fixed 9.

@ final: read: Post #50

Edited by adiace
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From what I understood... you start with max possible in 6digits then fixed 9, so its 9XXXXX and subtracted your calculation for 5 digits.. but when you do that, you must keep in mind the fixed 9.

@ final: read: Post #50

I did. That's why I surmised that any flippable numbers must end in 6 to be symmetrical.

so a number that contains all 0,1,6,8,9 must be 9xxxx6 to be flippable. so for any given 2nd digit, there are 5 possibilities, because the 5th digit must match, ie 90XX06, 91XX16, 96XX96, 98xx86, 99xx66. the middle two numbers must again match, so be either 00,11,69,88 or 96 5 possibilities for each.

Unless I'm missing something big my logic is sound.

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yah i thought you were wrong and then i realized your right glycerine (i think)

but man im spacy today i started doing all my math for the 30th day then... ok ignore all my latest posts (again)

Edited by final
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yah i thought you were wrong and then i realized your right glycerine (i think)

but man im spacy today i started doing all my math for the 30th day then... ok ignore all my latest posts (again)

We make a great team. Want to give up but can't. Can't focus. Make more work for ourselves. Eventually get it ;)

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@ G

You logic sounds good :)

I approached it from the other side and got a slightly different ans:

2^20 = 1048576

Possible numbers using 5 digits = 107328

1048576 - 107328 = 941248 // the number of telephone number still required.

9*10^5-4*5^5+4*5^2 = 887600 // all 6 digit numbers excluding 9xxxxx

5*10^4-2*5^4+2*5^2 = 48800 // all numbers starting with 96xxxx

4000 + 800 + 40 + 8 = 4848 // 95_______ numbers

941248 - 887600 - 48800 - 4848 = 0

the ans i got is 953737

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@ G

You logic sounds good :)

I approached it from the other side and got a slightly different ans:

2^20 = 1048576

Possible numbers using 5 digits = 107328

1048576 - 107328 = 941248 // the number of telephone number still required.

9*10^5-4*5^5+4*5^2 = 887600 // all 6 digit numbers excluding 9xxxxx

5*10^4-2*5^4+2*5^2 = 48800 // all numbers starting with 96xxxx

4000 + 800 + 40 + 8 = 4848 // 95_______ numbers

941248 - 887600 - 48800 - 4848 = 0

the ans i got is 953737

First glance nothing stands out which makes me wonder why we got different answers but I'll have to look more closely later. I'm at work and can't devote that kind of brain power atm :)

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i have stared at the problem so long that my own formulas no longer make sense

i reprogramed my java so much i cant fix it

my work papers are so messy i have no idea where any of them are

and i figured out i can do it relatively easily if i could just get it all to work again... so easily

so frustrating

im taking a sabbatical

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Hint:

Calculating the last telephone number will help calculate the "Sum of all telephone numbers on the 20th day"

@ final.. i have to admit, i had you in mind when i created this problem (specifically to make it very hard for you to write it into code)

Edited by adiace
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ok so i think i might have it or something close so here is the sum of all numbers (used and not) below 954887 (i looked at glycerines math and agree i looked at yours adiace and got confused so ill try to use my reasoning at your math on the bottom)

(477443*954887+477444)

+49999*100000+50000

+4999*10000+5000

+499*1000+500

+49*100+50

+4*10+5

(lets say the above equals x)

the excluded numbers between 0 and 1000000 sum equals 8432558304

the excluded numbers that need to be added back in because they are between 954887 and 1000000 are 1826518515

so i get

x-8432558304+1826518515

now eventually i might actually actually solve this but i dont know how a calculator chops stuff off and my computer seems to explode for some reason

anyway please solve that with you amazing spread sheet and get back to me about the horrible inaccuracies im sure i have

its flattering that you want to torture me more i wish i had friends like you in real life to destroy my brain

and to end this post

if im not wrong this is where you go wrong (i might just be misunderstanding what ur doing)

"5*10^4-2*5^4+2*5^2 = 48800 // all numbers starting with 96xxxx"

how can there be more then 10000 numbers starting with 96 this number should be 10^4-5^4+5

and glycerine this has been fun bouncing ideas with you even if half the time mine have been more confusion

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2^20 = 1048576

Possible numbers using 5 digits = 107328

1048576 - 107328 = 941248 // the number of telephone number still required.

9*10^5-4*5^5+4*5^2 = 887600 // all 6 digit numbers from 000000 to 899997

5*10^4-2*5^4+2*5^2 = 48800 // all 6 digit numbers starting from 900000 to 949999

4000 // all 6 digit numbers starting from 950000 to 953999

800 // all 6 digit numbers from 954000 to 954799

40 // all 6 digit numbers from 954800 to 954839

8 // all 6 digit numbers from 954840 to 954847

941248 - 887600 - 48800 - 4000 - 800 - 40 - 8 = 0

Corrected ans: 954847

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ok so i think i might have it or something close so here is the sum of all numbers (used and not) below 954887 (i looked at glycerines math and agree i looked at yours adiace and got confused so ill try to use my reasoning at your math on the bottom)

(477443*954887+477444)

+49999*100000+50000

+4999*10000+5000

+499*1000+500

+49*100+50

+4*10+5

(lets say the above equals x)

the excluded numbers between 0 and 1000000 sum equals 8432558304

the excluded numbers that need to be added back in because they are between 954887 and 1000000 are 1826518515

so i get

x-8432558304+1826518515

try http://instacalc.com/

i copy pasted your equation for x and got 460,955,040,880

x-8432558304+1826518515 = 454,349,001,091

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