Happy new year 2019 digit problem

BUT I was NOT looking for those: I was looking for 3 numbers that are
each 4 digits, NOTHING else. Which meant the lower was in range
1000 to 3333, meaning high case of 3333,6666,9999.
Sorry 'bout dat, chief...as Maxwell Smart would say...
Where's agent 99 when you need her?

-Dan
 
Give me the new "rules" plus an example...
Find numbers with the following properties: Find all numbers a s/t all digits will be found among a, 2a and 3a. Since the 3 numbers must contain all 10 digits, you can start with three digit numbers above 333. If you already started with 334, then just continue from where you left off at. Go as far as you reasonably can. It would be nice to have a 4th column that adds the digits in the 3rd column.
Example
1692 3384 5076 18
1809 3618 5427 18
1902 3804 5706 18

It would be extra nice if it can point out sums that are not multiples of 18 (or it is turns out to multiples of 9, then not multiples of 9)

Thanks!
 
I've already given you those.

I need an example where the 1st number is in range 334 to 999.
I'm baffled. Are you saying that your program needs an example to find other numbers??
Can you continue from where you left off from??
 
I'm baffled. Are you saying that your program needs an example to find other numbers??
Don't get ridiculous, Sir Jomo!
I am not finding any cases, so I wanted an example
(which I erroneously thought you had!)
in order to double-check my program.

So, as far as I'm concerned, there are NO cases
where the lowest number is less than 1000.

Prove me wrong: write your own program!!
 
Don't get ridiculous, Sir Jomo!
I am not finding any cases, so I wanted an example
(which I erroneously thought you had!)
in order to double-check my program.

So, as far as I'm concerned, there are NO cases
where the lowest number is less than 1000.

Prove me wrong: write your own program!!
Denis, I do not think that you understand my request. I never asked for numbers that were only three digits. Even my example had 10 digits. I would like you to please find all numbers that have the desired property. This includes 4 digit numbers, 5 digit numbers, etc. I understand that you can't find all, but go as far as you like.

Steve
 
Every "number of digits" change requires basically a
new program (I use UBasic); I picked 6 digits numbers.

Sum of digits of 3rd number is not steady; example:
100192 200384 300576 21
100458 200916 301374 18
100478 200956 301434 15

Anyhoo, there's 31657 cases, the 31657th being:
333270 666540 999810 36

That's it from me...
 
Every "number of digits" change requires basically a
new program (I use UBasic); I picked 6 digits numbers.

Sum of digits of 3rd number is not steady; example:
100192 200384 300576 21
100458 200916 301374 18
100478 200956 301434 15

Anyhoo, there's 31657 cases, the 31657th being:
333270 666540 999810 36

That's it from me...
That must have been a long walk......
 
… would like you to please find all numbers … I understand that you can't find all [numbers] …
Hoo boy.

Anyway, as the number of digits in the starting number increases, we'd reach a point where the starting number itself already contains the 10 digits. You want to keep going? ;)
 
Every "number of digits" change requires basically a
new program (I use UBasic); I picked 6 digits numbers.

Sum of digits of 3rd number is not steady; example:
100192 200384 300576 21
100458 200916 301374 18
100478 200956 301434 15

Anyhoo, there's 31657 cases, the 31657th being:
333270 666540 999810 36

That's it from me...
OK, you did good, thanks!
It seems that the sum of the digits of 3n is simply a multiple of 3 as expected. However we did not find a '3' yet, but I suspect it will be there.
 
Hoo boy.

Anyway, as the number of digits in the starting number increases, we'd reach a point where the starting number itself already contains the 10 digits. You want to keep going? ;)
I actually gave two examples where the 1st number (10 digits long) had all the digits and when multiplied by 3 the sum of the digits of this number was 36. I suspected that maybe the sum of the digits of 3n would be at a least a multiple of more than 3. It still may be true but I doubt it. I will run a program to see if I can get a 3. Of course the number 3n will either have three 1's and many 0's or a 2 and two 1's and many 0's. I guess there is a reason for programming after all (only joking).
 
I actually gave two examples where the 1st number (10 digits long) had all the digits and when multiplied by 3 the sum of the digits of this number was 36. I suspected that maybe the sum of the digits of 3n would be at a least a multiple of more than 3. It still may be true but I doubt it. I will run a program to see if I can get a 3. Of course the number 3n will either have three 1's and many 0's or a 2 and two 1's and many 0's. I guess there is a reason for programming after all (only joking).
Geezzzzzzz.....hope all that uselessness brings down the price of groceries :p
 
I actually gave two examples where the 1st number (10 digits long) had all the digits and when multiplied by 3 the sum of the digits of this number was 36. I suspected that maybe the sum of the digits of 3n would be at a least a multiple of more than 3. It still may be true but I doubt it. I will run a program to see if I can get a 3. Of course the number 3n will either have three 1's and many 0's or a 2 and two 1's and many 0's. I guess there is a reason for programming after all (only joking).


Geezzzzzzz.....hope all that uselessness brings down the price of groceries Tutuapp 9apps Showbox :p
Denis
i don't think the price of groceries will be donw :D
 
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