primes modulo 9

Question

I have an interesting property that I would like to share about odd prime numbers and I was wondering if anyone with a Number Theory background the reason why this happens.

I take a prime number and take multiples of that specific prime number. I then take the digits and add them and take them modulo 9. when looking at the list we see the cyclic pattern of the permutation group modulo 9. This only works for some primes but not others.

In example: 17 works

17, 34 , 51, 68, 85, 102, 119, 136, 153, ...

adding the digits:

8, 7, 6, 14, 13, 3, 11, 10, 9

taking modulo 9:

8,7,6,5,4,3,2,1,0.

Some are well ordered like 17, some are not but still cyclic in their pattern. I discussed with a college and he said it looked like the quadratic reciprocity property of the number that we are using modulo 9.

Answer 1

You're not really undertaking two steps here; you're just finding the value of each number generated $\bmod 9$, because taking a digit sum does not alter the $\bmod 9 $ value.

So, given that all primes except $3$ are coprime to $9$, their multiples will cycle through the possible congruence classes - as will any other number not a multiple of $3$, for example $20$.

Answer 2

Any number $n \equiv 8 \pmod 9$ will show that pattern you have observed, like $n = 8$ and $n = 26$, both of which are composite. See Sloane's A061242 for some more primes like $17$.

As you already know, $8 + 8 = 16$. And $16 = 7 \pmod 9$. Notice also $2 + 4 = 6$, $3 + 2 = 5$, $4 + 0 = 4$, $4 + 8 = 12$ and $1 + 2 = 3$, $5 + 6 = 11$ and $1 + 1 = 2$, $6 + 4 = 10$, and last but not least $7 + 2 = 9$.

Answer 3

Are you aware that $\sum \text{digits of } n \equiv n \mod 9$?

So if $k = 9j + i$ then $\sum \text{digits of }(9j+i)n \equiv (9j + i)n \equiv in \equiv \sum \text{digits of } in$.

So, yes, such operations on multiples of any number, prime or not, will have cyclic results of periods 9 or divisors of 9.

If $\gcd(9, n) = m$ then it will have cycles of period $9/m$ (because $(\frac 9mj + i)*n = (\frac 9mj + i)*m\frac nm = (9j + im)\frac nm \equiv im\frac nm \equiv in \mod 9$.

So if the $9|n$ then the cycle is $1$. Ex. $18,36, 54, 63$ yield $0,0,0,0$ if $3|n$ but $9 \not \mid n$ the cycle is $3$. Ex. $6,12,18,24, 30$ yields $6,3,0,6,3$. (Also note every value is a multiple of $3$). And if $\gcd(9,n) =1$ the cycle is $9$. Example: $14, 28, 42, 56, 70, 84, 98, 112, 126,140$ the cycle is $5,1,6,2,7,3,8, 4, 0, 5$.

Note: if $n \equiv 1 \mod 9$ the cycle will be from $1 $ to $8$ to $0$ in ascending order. $19, 38, 57, 76, 95, 114, 133,152, 171, 190$ yield $1,2,3,4,5,6,7,8,0,1$.

If $n \equiv 8 \equiv -1\mod 9$ the cycle be from $8$ to $0$ in descending order.

They will all be in the order $n, 2n \mod 9, 3n \mod 9, 4n \mod 9.....$.

Example: If we want a cycle that descends by 2 we do $n \equiv -2 \equiv 7 \mod 9$.

$16,32,48,64,80,96,112,128,144,160$ yields, $7,5,3,1,8,6,4,2,0,7$

And so on.

===addendum===

$$n = \sum a_i 10^k = \sum a_i *(99999..... + 1) = \sum 9999....a_i + \sum a_i \equiv \sum a_i \mod 9 = \sum \text{digits of } n \mod 9$$ (for the proof of the first line.)

Answer 4

I'm not a math whiz, but I have been looking into this. The patterns with primes is quite cool. The pattern with multiples of any numbers is quite cool.

If the digits in a number equal 2, their multiples' digit sums will follow the 2 pattern: 2,4,6,8,1,3,5,7,9........... repeating until the end of time. This is also true for all primes whose digits added together equal two.

If the digits in a number equal 3, they will follow the 3 pattern: 3,6,9,3,6,9,3,6,9,...........

For 4, the pattern is: 4,8,3,7,2,6,1,5,9,............

For 5, the pattern is: 5,1,6,2,7,3,8,4,9,..............

For 6, the pattern is: 6,3,9,6,3,9,6,3,9,................

For 7, the pattern is: 7,5,3,1,8,6,4,2,1,9,.................

For 8, the pattern is: 8,7,6,5,4,3,2,1,9,...................

For 9, the pattern is: 9,9,9,9,9,9,9,9,9,...................

The ninth digit in each pattern is always 9. The 4 and 5 patterns are mostly reversals of each other. The 8 and 1 patterns are mostly reversals of each other. The 3 and 6 patterns are similar but unique in that 3 and 6 are reversed. The 2 and 7 patterns are mostly reversals of each other. The 9 pattern is all nines.

I feel that someone will use this information one day to find the secret behind prime numbers. By reducing all primes to one digit numbers, you can find a relation between multiples of 9. For example: 2, 11, 29, 47, 83, 101, 137, 173, 191,.... all reduce down to the digit 2. Not only do they follow the 2 pattern above, they are also all prime numbers and are separated by some multiple of nine. 9, 18, 18, 36, 18, 36, 36,..... This is true for primes that reduce down to any common digit.

I don't know about you, but I find that amazing.

For an example of how both 2 and 191 follow the same pattern:

2=2 , DIGITSUM(191)=1+9+1=DIGITSUM(11)=1+1=2 4=4 , DIGITSUM(382)=3+8+2=DIGITSUM(13)=1+3=4 6=6 , DIGITSUM(573)=5+7+3=DIGITSUM(15)=1+5=6

Yes, I made up this digit sum idea, because I am not well versed in higher math concepts on an academic level. But I promise you, looking into these cool patterns can both bring you joy and drive you crazy.

I hope someone can reveal the secret to primes. God willing.

Blessings!