Sudoku numbers

I give a solved sudoku as an example:

6 3 2 1 5 7 4 9 8
1 7 4 9 3 8 5 2 6
5 8 9 4 2 6 1 3 7
3 6 1 8 7 9 2 4 5
8 2 7 5 4 1 9 6 3
9 4 5 2 6 3 7 8 1
4 9 3 7 8 5 6 1 2
7 1 8 6 9 2 3 5 4
2 5 6 3 1 4 8 7 9

One property of this square is that all figures of each horizontal line are different, and all figures of each vertical line are different (and all figures in every sub 3x3 square is different, but this is somewhat beyond my point for now).

Put every line together and read it as a number. (Example: The first horizontal line reads 632 million 157 thousand and 498, or just 632157498). Let's call this a sudoku number as a definition.

For fun I entered an arbitrary sudoku number into my prime identification program (a program that says if a number is a prime or not). The first horizontal soduko number is not a prime (Why?), the second was not a prime either (Why?), the third was not a prime either, but this is harder to see with your eyes only.

I went through all of the soduko numbers and found no primes at all.
A took the soduko numbers and reversed them, and I didn't found any.

My question is - Is there soduko numbers that are primes? Are they so rare that I didn't find any because I didn't search long enough? Please state a certain soduko prime or explain why there isn't any.

Originally posted by FabianFnas
I give a solved sudoku as an example:

6 3 2 1 5 7 4 9 8
1 7 4 9 3 8 5 2 6
5 8 9 4 2 6 1 3 7
3 6 1 8 7 9 2 4 5
8 2 7 5 4 1 9 6 3
9 4 5 2 6 3 7 8 1
4 9 3 7 8 5 6 1 2
7 1 8 6 9 2 3 5 4
2 5 6 3 1 4 8 7 9

One property of this square is that all figures of each horizontal line are different, and all figures of each vertical line are different (and ...[text shortened]... n't search long enough? Please state a certain soduko prime or explain why there isn't any.

No.

1+2+3+4+5+6+7+8+9=45, and 3 divides 45. Therefore, any 9 digit number that does not contain 0 is divisible by 3, and so any possible prime number that is 9 digits long and contains no number twice must contain a 0.

Also, an n by n grid that contains no number in any row or column twice is called a Latin Square. 🙂

Originally posted by Swlabr
No.

1+2+3+4+5+6+7+8+9=45, and 3 divides 45. Therefore, any 9 digit number that does not contain 0 is divisible by 3, and so any possible prime number that is 9 digits long and contains no number twice must contain a 0.

Also, an n by n grid that contains no number in any row or column twice is called a Latin Square. 🙂

A 9 digit number NOT divisible by 3.

111111112

EDIT: OK I see what you mean! (I dont do soduko) 😳

Originally posted by wolfgang59
A 9 digit number NOT divisible by 3.

111111112

That contains a number twice. You cannot make a prime number from 1,2,3,4,5,6,7,8,9 all used in one number.

so latin squares of size NxN cannot contain primes where N/3 is an integer.
also, they cannot contain primes where N/3 + 1/3 is an integer.

That would mean that 4x4, 7x7 can contain primes
(quick check: 4231, 1524763)

but are these the only two rules?

Originally posted by Swlabr
No.

1+2+3+4+5+6+7+8+9=45, and 3 divides 45. Therefore, any 9 digit number that does not contain 0 is divisible by 3, and so any possible prime number that is 9 digits long and contains no number twice must contain a 0.

Also, an n by n grid that contains no number in any row or column twice is called a Latin Square. 🙂

I went through all the trouble to add them up and all I had to do was scroll down 😛

Originally posted by pawnhandler
I went through all the trouble to add them up and all I had to do was scroll down 😛

Yes, it's a lot of trouble.

9+1
8+2
7+3
6+4
5

I didn't bother reading other peoples responses so somebody else could have said this.

if i understand you correctly you are either talking about the entire soduko board or just a colum. if so then it will never be a prime because it will always have the same digits; 1,2,3,4,5,6,7,8 and 9 in each row.

Originally posted by Smider
I didn't bother reading other peoples responses so somebody else could have said this.

if i understand you correctly you are either talking about the entire soduko board or just a colum. if so then it will never be a prime because it will always have the same digits; 1,2,3,4,5,6,7,8 and 9 in each row.

Only one row of digits, or column.

You say "...it will never be a prime because it will always have the same digits; 1,2,3,4,5,6,7,8 and 9 in each row"

Only the reason that two numbers has the same digits in them doesn't say that one sequence of the numbers is prime or not compared with another sequense of the same digits.

Example: 23 is prime, 32 is not.

But, as said earlier, if a number is dividable with 3, the sum of all digits included is dividable with 3, no matter what sequence.

So the common property with all sudoku-numbers is that they are all dividable with three, we are all agreed to that (?). Is there another property as well, worth mentioning? (Yes, there is. Which?)

Originally posted by FabianFnas
Only one row of digits, or column.

You say "...it will never be a prime because it will always have the same digits; 1,2,3,4,5,6,7,8 and 9 in each row"

Only the reason that two numbers has the same digits in them doesn't say that one sequence of the numbers is prime or not compared with another sequense of the same digits.

Example: 23 is prime, ...[text shortened]... eed to that (?). Is there another property as well, worth mentioning? (Yes, there is. Which?)

The divisibility for three is enough

Originally posted by adam warlock
The divisibility for three is enough

Is there any other multiplier all sudoku numbers have in common?

Originally posted by FabianFnas
Is there any other multiplier all sudoku numbers have in common?

I tried to prof both answers but couldn't so I have to say I don't know but I'm more inclined to say that there isn't.

they also divide by 9, though I wouldn't bother testing that experimentally.

Originally posted by doodinthemood
they also divide by 9, though I wouldn't bother testing that experimentally.

Yezz. Now I am satisfied.

Test for divisibility for three - If the sum of the figures is divisable with 3 then the number is also divisable with three. 1+2+3+4+5+6+7+8+9=45, 45 is divisable with 3.

Test for divisability for nine - If the sum of the figures is divisable with 9 then the number is also divisable with nine. 1+2+3+4+5+6+7+8+9=45, 45 is divisable with 9.

Originally posted by FabianFnas
Yezz. Now I am satisfied.

Test for divisibility for three - If the sum of the figures is divisable with 3 then the number is also divisable with three. 1+2+3+4+5+6+7+8+9=45, 45 is divisable with 3.

Test for divisability for nine - If the sum of the figures is divisable with 9 then the number is also divisable with nine. 1+2+3+4+5+6+7+8+9=45, 45 is divisable with 9.

I guess I missinterpreted your question. When is interested in divisibility one usually speaks of factorisation in prime numbers. So I was trying to deliver a proof with prime factorisation.