/sci/ - Science & Math

Here is ma partial result. I assume the expected solution uses modular arithmetic as a tool, which I don't know much about.

Let <span class="math">b[/spoiler] be the base (for exmaple 10 in your pic.)
The number with all the successive digits, i.e. the string of the form
<span class="math">123\dots (b-2)(b-1)[/spoiler]
(for ample 123456789)
can be written as
<span class="math">\sum_{n=0}^{b-1}(b-1-n)\cdot b^n[/spoiler].
For example
<span class="math">\sum_{n=0}^{9}(9-n)\cdot 10^n[/spoiler].

Now you multiply this by a number <span class="math">r<span class="math"> with 0<r<b.
The last digit of
r\times\sum_{n=0}^{b-1}(b-1-n)\cdot b^n
is
r\times (b-1-n)\ \mathrm{mod}
with n=0, i.e.
r\times (b-1-0)\ \mathrm{mod}=(r-1)b+(b-r)\ \mathrm{mod}=b-r.
This is clear and you see this in your picture: When the number gets multiplied by 2, the last digit of the result is 10-2=8 and when you multiply it be 6, it's similarly 10-6=4.
Now from (r-1)b+(b-r)\ \mathrm{mod}, the number (r-1) gets passed on to the next numer digit at n=1:
(r\times (b-1-1))+(r-1)\ \mathrm{mod}=(r-2)b+(b-(r+1))\ \mathrm{mod}=(b-r)-1.
This is why the second to last digit is always the last digit minus one.
Now I think I don't need to make the next calculation explicitly: From the picture it looks like the third to last digit is the second to last digit minus r.
For example, the r=4 case has as last digit b-r=6, then b-r-1=5, then b-r-r=5.
You could calculate this that way, but it feels a lit lenghty and there is certainly a more sophisticated way to do it.[/spoiler][/spoiler]

Here is ma partial result. I assume the expected solution uses modular arithmetic as a tool, which I don't know much about.

Let <span class="math">b[/spoiler] be the base (for exmaple 10 in your pic.)
The number with all the successive digits, i.e. the string of the form
<span class="math">123\dots (b-2)(b-1)[/spoiler]
(for ample 123456789)
can be written as
<span class="math">\sum_{n=0}^{b-1}(b-1-n)\cdot b^n[/spoiler].
For example
<span class="math">\sum_{n=0}^{9}(9-n)\cdot 10^n[/spoiler].

Now you multiply this by a number r with <span class="math">0<r<b[/spoiler].
The last digit of
<span class="math">r\times\sum_{n=0}^{b-1}(b-1-n)\cdot b^n[/spoiler]
is
<span class="math">r\times (b-1-n)\ \mathrm{mod}[/spoiler]
with <span class="math">n=0[/spoiler], i.e.
<span class="math">r\times (b-1-0)\ \mathrm{mod}=(r-1)b+(b-r)\ \mathrm{mod}=b-r[/spoiler].
This is clear and you see this in your picture: When the number gets multiplied by 2, the last digit of the result is 10-2=8 and when you multiply it be 6, it's similarly 10-6=4.
Now from <span class="math">(r-1)b+(b-r)\ \mathrm{mod}[/spoiler], the number <span class="math">(r-1)[/spoiler] gets passed on to the next numer digit at n=1:
<span class="math">(r\times (b-1-1))+(r-1)\ \mathrm{mod}=(r-2)b+(b-(r+1))\ \mathrm{mod}=(b-r)-1[/spoiler].
This is why the second to last digit is always the last digit minus one.
Now I think I don't need to make the next calculation explicitly: From the picture it looks like the third to last digit is the second to last digit minus r.
For example, the r=4 case has as last digit b-r=6, then b-r-1=5, then b-r-r=5.
You could calculate this that way, but it feels a lit lenghty and there is certainly a more sophisticated way to do it.
The bottom line is this:
To say that all numbers appear in the string of b-1 symbols is to say that no number appears twice. The picture suggest that it fails if a 0 appears.
And in the cases 3,6,9, it's always the case that first 0 shows up and then b-r, the last digit, repeats.
I guess the answer will be that it's always this pattern and also 3,6,9 not working might mean that multiples of factors of b-1 don't work.
I hope this gives ideas.

>>5928876
I'm a Java fag.

Let me write you a program and see if we can identify a pattern. It looks like it may be multiples of three, but I'm curious.

>>5929228
>Java

kill yourself

"mod" denotes "mod b" of course

>>5929229
>>5929228
123456789 * 1 = 123456789 all 9: true
123456789 * 2 = 246913578 all 9: true
123456789 * 3 = 370370367 all 9: false
123456789 * 4 = 493827156 all 9: true
123456789 * 5 = 617283945 all 9: true
123456789 * 6 = 740740734 all 9: false
123456789 * 7 = 864197523 all 9: true
123456789 * 8 = 987654312 all 9: true
123456789 * 9 = 1111111101 all 9: false
123456789 * 10 = 1234567890 all 9: true
123456789 * 11 = 1358024679 all 9: true
123456789 * 12 = 1481481468 all 9: false
123456789 * 13 = 1604938257 all 9: true
123456789 * 14 = 1728395046 all 9: true
123456789 * 15 = 1851851835 all 9: false
123456789 * 16 = 1975308624 all 9: true
123456789 * 17 = 2098765413 all 9: true
123456789 * 18 = 2222222202 all 9: false
123456789 * 19 = 2345678991 all 9: true
123456789 * 20 = 2469135780 all 9: true
123456789 * 21 = 2592592569 all 9: false
123456789 * 22 = 2716049358 all 9: true
123456789 * 23 = 2839506147 all 9: true
123456789 * 24 = 2962962936 all 9: false
123456789 * 25 = 3086419725 all 9: true
123456789 * 26 = 3209876514 all 9: true
123456789 * 27 = 3333333303 all 9: false
123456789 * 28 = 3456790092 all 9: false
123456789 * 29 = 3580246881 all 9: false
123456789 * 30 = 3703703670 all 9: false
123456789 * 31 = 3827160459 all 9: true
123456789 * 32 = 3950617248 all 9: true
123456789 * 33 = 4074074037 all 9: false
123456789 * 34 = 4197530826 all 9: true
123456789 * 35 = 4320987615 all 9: true
123456789 * 36 = 4444444404 all 9: false
123456789 * 37 = 4567901193 all 9: false
123456789 * 38 = 4691357982 all 9: true
123456789 * 39 = 4814814771 all 9: false
123456789 * 40 = 4938271560 all 9: true
123456789 * 41 = 5061728349 all 9: true
123456789 * 42 = 5185185138 all 9: false
123456789 * 43 = 5308641927 all 9: true
123456789 * 44 = 5432098716 all 9: true
123456789 * 45 = 5555555505 all 9: false
123456789 * 46 = 5679012294 all 9: false

>>5929251
123456789 * 47 = 5802469083 all 9: false
123456789 * 48 = 5925925872 all 9: false
123456789 * 49 = 6049382661 all 9: false
123456789 * 50 = 6172839450 all 9: true
123456789 * 51 = 6296296239 all 9: false
123456789 * 52 = 6419753028 all 9: true
123456789 * 53 = 6543209817 all 9: true
123456789 * 54 = 6666666606 all 9: false
123456789 * 55 = 6790123395 all 9: false
123456789 * 56 = 6913580184 all 9: false
123456789 * 57 = 7037036973 all 9: false
123456789 * 58 = 7160493762 all 9: false
123456789 * 59 = 7283950551 all 9: false
123456789 * 60 = 7407407340 all 9: false
123456789 * 61 = 7530864129 all 9: true
123456789 * 62 = 7654320918 all 9: true
123456789 * 63 = 7777777707 all 9: false
123456789 * 64 = 7901234496 all 9: false
123456789 * 65 = 8024691285 all 9: false
123456789 * 66 = 8148148074 all 9: false
123456789 * 67 = 8271604863 all 9: false
123456789 * 68 = 8395061652 all 9: false
123456789 * 69 = 8518518441 all 9: false
123456789 * 70 = 8641975230 all 9: true
123456789 * 71 = 8765432019 all 9: true
123456789 * 72 = 8888888808 all 9: false
123456789 * 73 = 9012345597 all 9: false
123456789 * 74 = 9135802386 all 9: false
123456789 * 75 = 9259259175 all 9: false
123456789 * 76 = 9382715964 all 9: true
123456789 * 77 = 9506172753 all 9: false
123456789 * 78 = 9629629542 all 9: false
123456789 * 79 = 9753086331 all 9: false
123456789 * 80 = 9876543120 all 9: true
123456789 * 81 = 9999999909 all 9: false
123456789 * 82 = 10123456698 all 9: false
123456789 * 83 = 10246913487 all 9: false
123456789 * 84 = 10370370276 all 9: false
123456789 * 85 = 10493827065 all 9: true
123456789 * 86 = 10617283854 all 9: false
123456789 * 87 = 10740740643 all 9: false
123456789 * 88 = 10864197432 all 9: false
123456789 * 89 = 10987654221 all 9: false
123456789 * 90 = 11111111010 all 9: false
123456789 * 91 = 11234567799 all 9: false
123456789 * 92 = 11358024588 all 9: false
123456789 * 93 = 11481481377 all 9: false

>>5929251
Good effort, but the focus should be on different base systems, not on multiplications by numbers beyond the digits of that base.

>>5928876
Just in case :
a number which has this property is called a , (zeroless) pandigital number.

You'll find a lot of papers on google and examples on oeis

>>5929254
123456789 * 94 = 11604938166 all 9: false
123456789 * 95 = 11728394955 all 9: false
123456789 * 96 = 11851851744 all 9: false
123456789 * 97 = 11975308533 all 9: false
123456789 * 98 = 12098765322 all 9: false
123456789 * 99 = 12222222111 all 9: false

>>5929251
>>5929254
What impressive skills !!
I hope you didn't broke your computer

Java for this, u serious ?
I can do this under DOS -_-

>>5929260
bro. calm down.

I'm just having fun.

>>5929263
I'm calm :) I'm having fun also seeing that.

Don't take it badly.
But use pastebin or zerobin if you want to paste more, it's going to be unreadeable.

>>5929266
yeah should have, but i wanted to make a big post haha.

like my third time on sci. i'm just enjoying life trying to chill so i'm not really in the swing of things.

>>5928876
seems to hold true with hex from very limited testing

bump for interest

OP What you think about this?

>>5929260
>>5929229
>java is shit
>I can do stuff in basic too that you do in java
What is your problem seriously, I can do this shit in assembler too if that satisfies your superiority complex. The guy did it in java, who cares.

>>5931130
The only way a language will die is if the programmers of it are endlessly ridiculed for writing in it.

Maybe it's something like this

For each integer x where 1/x is a repeating decimal, all multiples of x when multiplied by 123456789 will not produce a pandigital number.

>>5931130
>learning second degree

USfag ? You never learn irony in your country ? :)

>>5931101

Should we abolish 7?

>>5931101
I guess that is what happens when you combine yellows with reds. (a prime number that is the sum of two other primes? idk)

>>5932251
what u mean?
7 is a great number see what we get
142857 in period!!!
its a secret number!

>>5932263
>I guess that is what happens when you combine yellows with reds. (a prime number that is the sum of two other primes? idk)

did not understand the idea.

>>5928876
Hey OP
turns out when you multiply numbers like that, by the highest digit, you get a result of 111...1101 regardless of which base you're using. give it a try on base 3, for example

>>5929226
only decent post of the thread gets ignored. bye /sci/

solved?

one last call for a solution

>>5932292
see >>5929256
looks to me like the pot calling the kettle black

>>5934610
it seems the pandigital numbers don't mind other number appear. Here it is about a rearanging of the string 1234...