The multiple of nine mystery
 Mysterio448
 Posts: 368
 Joined: May 3rd, 2013, 6:44 pm
The multiple of nine mystery
There is a strange mathematical pattern that has come to my attention. I don't know how many people have noticed this, but for any multiple of nine the sum of the digits in that number equals nine. Here is what I mean:
9*1=9
9*2=18, 1+8=9
9*3=27, 2+7=9
9*4=36, 3+6=9
9*5=45, 4+5=9
9*6=54, 5+4=9
9*7=63, 6+3=9
9*8=72, 7+2=9
9*9=81, 8+1=9
9*10=90, 9+0=9
9*11=99, 9+9=18, 1+8=9
9*12=108, 1+0+8=9
9*13=117, 1+1+7=9
9*14=126, 1+2+6=9
9*15=135, 1+3+5=9
9*16=144, 1+4+4=9
9*17=153, 1+5+3=9
9*18=162, 1+6+2=9
9*19=171, 1+7+1=9
. . . and so on.
My question is: Why does this happen? What is the logical or mathematical explanation for this strange pattern?
9*1=9
9*2=18, 1+8=9
9*3=27, 2+7=9
9*4=36, 3+6=9
9*5=45, 4+5=9
9*6=54, 5+4=9
9*7=63, 6+3=9
9*8=72, 7+2=9
9*9=81, 8+1=9
9*10=90, 9+0=9
9*11=99, 9+9=18, 1+8=9
9*12=108, 1+0+8=9
9*13=117, 1+1+7=9
9*14=126, 1+2+6=9
9*15=135, 1+3+5=9
9*16=144, 1+4+4=9
9*17=153, 1+5+3=9
9*18=162, 1+6+2=9
9*19=171, 1+7+1=9
. . . and so on.
My question is: Why does this happen? What is the logical or mathematical explanation for this strange pattern?
 Spiral Out
 Site Admin
 Posts: 5007
 Joined: June 26th, 2012, 10:22 am
Re: The multiple of nine mystery
Yeah I noticed that. I play around with numbers sometimes as well and noticed this phenomenon too.
Another interesting quality... when adding any string of numbers (as in number reduction), all of the 9s (or numbers adding up to 9) can be removed without affecting the sum.
Example:
3+7+12+27+5+43+19+9+2+6+54=187 / 1+8+7=16 / 1+6=7  or just remove the 9 (the 1 & 8 from 187)
Remove all the 9s from the string above:
3&6, 7&2, 9
We are left with:
12+27+5+43+19+54=160 / 1+6+0=7
We can further remove the other 9s:
27 (2&7=9) and 54 (5&4=9)
We are then left with:
12+5+43+19=79 / 7+9=16 / 1+6=7 (or just remove the 9 from 79)
We can even remove the 9 from the 19:
12+5+43+1=61 / 6+1=7
The nines don't seem to matter!
Another interesting quality... when adding any string of numbers (as in number reduction), all of the 9s (or numbers adding up to 9) can be removed without affecting the sum.
Example:
3+7+12+27+5+43+19+9+2+6+54=187 / 1+8+7=16 / 1+6=7  or just remove the 9 (the 1 & 8 from 187)
Remove all the 9s from the string above:
3&6, 7&2, 9
We are left with:
12+27+5+43+19+54=160 / 1+6+0=7
We can further remove the other 9s:
27 (2&7=9) and 54 (5&4=9)
We are then left with:
12+5+43+19=79 / 7+9=16 / 1+6=7 (or just remove the 9 from 79)
We can even remove the 9 from the 19:
12+5+43+1=61 / 6+1=7
The nines don't seem to matter!
Dedicated to the fine art of thinking.

 Posts: 885
 Joined: May 9th, 2012, 8:05 am
 Location: The Evening Star
Re: The multiple of nine mystery
I think the clue as to why it's true is to note that it's a rule that can be generalized to any number base. For number base N, the recursively calculated sum of the digits of any multiple of the number N1 is equal to N1. So it's because 9 is one less than the number base we're using (10). It's a property of the way that number base systems work.
For example, hexadecimal (Base 16. Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F):
F * 1 = F
F * 2 = 1E, 1 + E = F
F * 3 = 2D, 2 + D = F
F * 4 = 3C, 3 + C = F
.
.
F * 10 = F 0, F + 0 = F
etc
Binary (Base 2. Digits: 0, 1):
1 * 1 = 1
1 * 10 = 10, 1 + 0 = 1
1 * 11 = 11, 1 + 1 = 10, 1 + 0 = 1
1 * 100 = 100, 1 + 0 + 0 = 1
1 * 101 = 101, 1 + 0 + 1 = 10, 1 + 0 = 1
etc
Each time you add 1 to the multiplier on the right of the '*' operator you are adding N1 to the sum. Because the number base is N, for any number not ending in 0, you are subtracting 1 from the last digit and, by carrying, adding 1 to the number which forms everything except the last digit. So their sum remains constant. When the number which forms everything except the last digit reaches a multiple of N1, the last digit has always fallen to 0. So the next addition of N1, on this occasion, doesn't cause a carry from the last digit. It simply flips the last digit back up to N1. i.e. it adds N1 to the sum.
That sounded a lot clearer in my head than it looks on the page.
For example, hexadecimal (Base 16. Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F):
F * 1 = F
F * 2 = 1E, 1 + E = F
F * 3 = 2D, 2 + D = F
F * 4 = 3C, 3 + C = F
.
.
F * 10 = F 0, F + 0 = F
etc
Binary (Base 2. Digits: 0, 1):
1 * 1 = 1
1 * 10 = 10, 1 + 0 = 1
1 * 11 = 11, 1 + 1 = 10, 1 + 0 = 1
1 * 100 = 100, 1 + 0 + 0 = 1
1 * 101 = 101, 1 + 0 + 1 = 10, 1 + 0 = 1
etc
Each time you add 1 to the multiplier on the right of the '*' operator you are adding N1 to the sum. Because the number base is N, for any number not ending in 0, you are subtracting 1 from the last digit and, by carrying, adding 1 to the number which forms everything except the last digit. So their sum remains constant. When the number which forms everything except the last digit reaches a multiple of N1, the last digit has always fallen to 0. So the next addition of N1, on this occasion, doesn't cause a carry from the last digit. It simply flips the last digit back up to N1. i.e. it adds N1 to the sum.
That sounded a lot clearer in my head than it looks on the page.
 LuckyR
 Posts: 2512
 Joined: January 18th, 2015, 1:16 am
Re: The multiple of nine mystery
Does this hold true in base 6?Mysterio448 wrote:There is a strange mathematical pattern that has come to my attention. I don't know how many people have noticed this, but for any multiple of nine the sum of the digits in that number equals nine. Here is what I mean:
9*1=9
9*2=18, 1+8=9
9*3=27, 2+7=9
9*4=36, 3+6=9
9*5=45, 4+5=9
9*6=54, 5+4=9
9*7=63, 6+3=9
9*8=72, 7+2=9
9*9=81, 8+1=9
9*10=90, 9+0=9
9*11=99, 9+9=18, 1+8=9
9*12=108, 1+0+8=9
9*13=117, 1+1+7=9
9*14=126, 1+2+6=9
9*15=135, 1+3+5=9
9*16=144, 1+4+4=9
9*17=153, 1+5+3=9
9*18=162, 1+6+2=9
9*19=171, 1+7+1=9
. . . and so on.
My question is: Why does this happen? What is the logical or mathematical explanation for this strange pattern?
"As usual... it depends."

 Posts: 885
 Joined: May 9th, 2012, 8:05 am
 Location: The Evening Star
Re: The multiple of nine mystery
Yes. Try it and see.
 Renee
 Posts: 327
 Joined: May 3rd, 2015, 10:39 pm
 Favorite Philosopher: Frigyes Karinthy
Re: The multiple of nine mystery
We were taught in Grade 7 in Hungary that you can check for divisibility by nine or by three by adding the digits, reductiowise, and if the last reduction results in 3 or 9, or in a number that is divisible by 3 or 9, then the original (nonreduced) number is divisible by 3 or nine.
I suspect it works for any positive integer power of three. (1, 3, 9, 27, 3^4, 3^5, ... 3^n where n is a positive integer.)
I don't know why this works. Nine is not mysterious. There is a proof somewhere. But don't ask "why" as if there were a divine decree about numbers.
Actually, I am wrong. The real reason this works is because Pope Poncificatius XIV in the eighth century Holy Roman Empire decreed it in his inaugural Easter sermon.
 Updated November 15th, 2016, 8:29 am to add the following 
Plus, you are wrong. (Sorry.)
In base6,
nine=13
and an example of 9*4=36 3+6=9 does not work in base six;
13*4=100 and 1+0+0 does not equal 13.
I suspect it works for any positive integer power of three. (1, 3, 9, 27, 3^4, 3^5, ... 3^n where n is a positive integer.)
I don't know why this works. Nine is not mysterious. There is a proof somewhere. But don't ask "why" as if there were a divine decree about numbers.
Actually, I am wrong. The real reason this works is because Pope Poncificatius XIV in the eighth century Holy Roman Empire decreed it in his inaugural Easter sermon.
 Updated November 15th, 2016, 8:29 am to add the following 
LuckyR wrote:Does this hold true in base 6?
That was a trick question. There is no digit "9" in a basesix number system.Dolphin42 wrote:Yes. Try it and see.
Plus, you are wrong. (Sorry.)
In base6,
nine=13
and an example of 9*4=36 3+6=9 does not work in base six;
13*4=100 and 1+0+0 does not equal 13.
Ignorance is power.

 Posts: 885
 Joined: May 9th, 2012, 8:05 am
 Location: The Evening Star
Re: The multiple of nine mystery
Renee:
As an example (in base 6):
5 * 4 = 32, 3 + 2 = 5
Try another one.
See my post #3. In base 6, the relevant number is 5. In base N the relevant number is N1.That was a trick question. There is no digit "9" in a basesix number system.
As an example (in base 6):
5 * 4 = 32, 3 + 2 = 5
Try another one.
 Spiral Out
 Site Admin
 Posts: 5007
 Joined: June 26th, 2012, 10:22 am
Re: The multiple of nine mystery
5 * 5 = 25, 2 + 5 = 7Dolphin42 wrote:Renee:See my post #3. In base 6, the relevant number is 5. In base N the relevant number is N1.That was a trick question. There is no digit "9" in a basesix number system.
As an example (in base 6):
5 * 4 = 32, 3 + 2 = 5
Try another one.
5 * 6 = 30, 3 + 0 = 3
Doesn't seem to work.
Dedicated to the fine art of thinking.
 Renee
 Posts: 327
 Joined: May 3rd, 2015, 10:39 pm
 Favorite Philosopher: Frigyes Karinthy
Re: The multiple of nine mystery
Dolphin42 wrote: As an example (in base 6):
5 * 4 = 32, 3 + 2 = 5
Try another one.
Dolphin is actually right, Spiral.Spiral Out wrote:5 * 5 = 25, 2 + 5 = 7
5 * 6 = 30, 3 + 0 = 3
Doesn't seem to work.
5*5 in base six is 41. 4+1 = 5.
Ignorance is power.
 Philosophy Explorer
 Posts: 2116
 Joined: May 25th, 2013, 8:41 pm
Re: The multiple of nine mystery
Here's another twist:
First I define anagram numbers (note the plural) by using a specific threedigit number. If we have abc where a, b and c take on a specific value from 0 to 9 and rearrange the digits to say cab, then take the difference so we have abc  cab = r, then r, the result, is always evenly divisible by 9 (i.e. no remainder) and abc and cab are anagram numbers.
To extend this further, as long as the rearranged numbers have the same number of digits (the anagram numbers), then r will always be evenly divisible by nine.
PhilX
First I define anagram numbers (note the plural) by using a specific threedigit number. If we have abc where a, b and c take on a specific value from 0 to 9 and rearrange the digits to say cab, then take the difference so we have abc  cab = r, then r, the result, is always evenly divisible by 9 (i.e. no remainder) and abc and cab are anagram numbers.
To extend this further, as long as the rearranged numbers have the same number of digits (the anagram numbers), then r will always be evenly divisible by nine.
PhilX
 Spiral Out
 Site Admin
 Posts: 5007
 Joined: June 26th, 2012, 10:22 am
Re: The multiple of nine mystery
Thanks for the clarification!Renee wrote:Dolphin is actually right, Spiral.
5*5 in base six is 41. 4+1 = 5.
Dedicated to the fine art of thinking.
 Philosophy Explorer
 Posts: 2116
 Joined: May 25th, 2013, 8:41 pm
Re: The multiple of nine mystery
Let's take this to the next level.Philosophy Explorer wrote:Here's another twist:
First I define anagram numbers (note the plural) by using a specific threedigit number. If we have abc where a, b and c take on a specific value from 0 to 9 and rearrange the digits to say cab, then take the difference so we have abc  cab = r, then r, the result, is always evenly divisible by 9 (i.e. no remainder) and abc and cab are anagram numbers.
To extend this further, as long as the rearranged numbers have the same number of digits (the anagram numbers), then r will always be evenly divisible by nine.
PhilX
As explained above, I've already said that abc  cab = r is (always) evenly divisible by nine. Now how about (abc)^{n}  (cab)^{n} = r_{1} where n is a natural number? I still maintain that r_{1} is evenly divisible by nine, furthermore it doesn't matter how many digits are involved in the bases (as long as they're anagram numbers) and I challenge anyone to find an exception. Test it out on your computers and scientific calculators.
PhilX
 Updated December 6th, 2016, 1:40 am to add the following 
I've checked my notes and I see this phenomenon extends to the binomial equation:
(x  y)^{n}  (x^{n}  y^{n}) = remainder
It would be helpful to consider a more specific example:
(x  y)^{3}  (x^{3}  y^{3}) =
3x^{2}y + 3xy^{2}
The first term has already been covered by my first post on this thread and is completely proven to work with any anagram numbers. The second term is covered in my second post and is partially proven to be compatible with any anagram numbers, but I'm completely confident it'll work for all anagram numbers. The third term, on the right side of the equal sign, I'm also completely confident is evenly divisible by nine so it can be represented by any anagram numbers such as 213 and 231 or 348 and 483 or 1225 and 5212, etc.
Since we're talking about the binomial equation, then n can be any natural number and x and y are anagram numbers (but the algebra can be a bit messy).
PhilX

 Posts: 28
 Joined: December 15th, 2016, 7:36 am
Re: The multiple of nine mystery
In the following, if a and b are integers, then a  b means that b is divisible by a. Also, ℕ is the set of (positive) natural numbers, ℕ = {1, 2, 3, 4, …}, and ℕ_0 is equal to {0, 1, 2, 3, …} = ℕ ∪ {0}.
Theorem: ∀ N, d, k, r, a_0, ..., a_r ∈ ℕ_0 :
((N ≥ 2) ∧ (d ≥ 2)∧ (k ≥ 1) ∧ (d^k = N – 1) ∧ (0 ≤ a_0, ..., a_r ≤ N – 1) ⇒
(d  (a_r * N^r + … + a_1 * N^1 + a_0 * N^0) ⇔ d  (a_r + … + a_1 + a_0)))
Proof: Let N, d, k, r, a_0, ..., a_r ∈ ℕ_0 be such that N ≥ 2 and d ≥ 2 and k ≥ 1 and d^k = N – 1 and 0 ≤ a_0, ..., a_r ≤ N – 1 and otherwise arbitrary.
For all j ∈ ℕ,
d * (d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (d^k) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (N – 1) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (N – 1) * N^(j – 1) + … + (N – 1) * N^2 + (N – 1) * N + (N – 1) + 1
= (N – 1) * N^(j – 1) + … + (N – 1) * N^2 + (N – 1) * N + N
= N * ((N – 1) * N^(j – 2) + … + (N – 1) * N + (N – 1) + 1)
= N * ((N – 1) * N^(j – 2) + … + (N – 1) * N + N)
= N^2 * ((N – 1) * N^(j – 3) + … + (N – 1) + 1)
= …
= N^(j – 1) * ((N – 1) * N^(j – j) + 1)
= N^(j – 1) * ((N – 1) * 1 + 1)
= N^(j – 1) * N
= N^j,
from which it follows that
(N^j) mod d
= (d * (d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1) + 1) mod d
= (d mod d) * (((d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1)) mod d) + (1 mod d)
= 0 + 1 because d mod d = 0 and 1 mod d = 1 due to d > 1,
= 1 (equation 1).
Thus, we obtain the following:
d  (a_r * N^r + … + a_1 * N^1 + a_0 * N^0)
⇔ 0 = (a_r * N^r + … + a_1 * N + a_0 * 1) mod d
= ((a_r * N^r) mod d) + … + ((a_1 * N) mod d) + (a_0 mod d)
= (a_r mod d)*(N^r mod d) + … + (a_1 mod d)*(N mod d) + (a_0 mod d)
= (a_r mod d)*1 + … + (a_1 mod d)*1 + (a_0 mod d) because of equation (1),
= (a_r + … + a_1 + a_0) mod d
⇔ d  (a_r + … + a_1 + a_0). Q.E.D
If we substitute N = 10, d = 9 and k = 1 in the above theorem, we find that a natural number m = a_r * 10^r + … + a_1 * 10 + a_0 is divisible by 9 if and only if the sum of its digits a_r + … + a_1 + a_0 is divisible by 9.
If we substitute N = 10, d = 3 and k = 2 in the above theorem, we find that a natural number m = a_r * 10^r + … + a_1 * 10 + a_0 is divisible by 3 if and only if the sum of its digits a_r + … + a_1 + a_0 is divisible by 3.
As can be seen, there is nothing mysterious about this quite useful result, which we have shown to be true for every base greater than or equal to 2 and every number that has a power equal to that base – 1.
 Updated January 10th, 2017, 5:39 pm to add the following 
Theorem: ∀ N, d, k, r, a_0, ..., a_r ∈ ℕ_0 :
((N ≥ 2) ∧ (d ≥ 2)∧ (k ≥ 1) ∧ (d^k = N – 1) ∧ (0 ≤ a_0, ..., a_r ≤ N – 1) ⇒
(d  (a_r * N^r + … + a_1 * N^1 + a_0 * N^0) ⇔ d  (a_r + … + a_1 + a_0)))
Proof: Let N, d, k, r, a_0, ..., a_r ∈ ℕ_0 be such that N ≥ 2 and d ≥ 2 and k ≥ 1 and d^k = N – 1 and 0 ≤ a_0, ..., a_r ≤ N – 1 and otherwise arbitrary.
For all j ∈ ℕ,
d * (d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (d^k) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (N – 1) * (N^(j – 1) + … + N^2 + N + 1) + 1
= (N – 1) * N^(j – 1) + … + (N – 1) * N^2 + (N – 1) * N + (N – 1) + 1
= (N – 1) * N^(j – 1) + … + (N – 1) * N^2 + (N – 1) * N + N
= N * ((N – 1) * N^(j – 2) + … + (N – 1) * N + (N – 1) + 1)
= N * ((N – 1) * N^(j – 2) + … + (N – 1) * N + N)
= N^2 * ((N – 1) * N^(j – 3) + … + (N – 1) + 1)
= …
= N^(j – 1) * ((N – 1) * N^(j – j) + 1)
= N^(j – 1) * ((N – 1) * 1 + 1)
= N^(j – 1) * N
= N^j,
from which it follows that
(N^j) mod d
= (d * (d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1) + 1) mod d
= (d mod d) * (((d^(k – 1)) * (N^(j – 1) + … + N^2 + N + 1)) mod d) + (1 mod d)
= 0 + 1 because d mod d = 0 and 1 mod d = 1 due to d > 1,
= 1 (equation 1).
Thus, we obtain the following:
d  (a_r * N^r + … + a_1 * N^1 + a_0 * N^0)
⇔ 0 = (a_r * N^r + … + a_1 * N + a_0 * 1) mod d
= ((a_r * N^r) mod d) + … + ((a_1 * N) mod d) + (a_0 mod d)
= (a_r mod d)*(N^r mod d) + … + (a_1 mod d)*(N mod d) + (a_0 mod d)
= (a_r mod d)*1 + … + (a_1 mod d)*1 + (a_0 mod d) because of equation (1),
= (a_r + … + a_1 + a_0) mod d
⇔ d  (a_r + … + a_1 + a_0). Q.E.D
If we substitute N = 10, d = 9 and k = 1 in the above theorem, we find that a natural number m = a_r * 10^r + … + a_1 * 10 + a_0 is divisible by 9 if and only if the sum of its digits a_r + … + a_1 + a_0 is divisible by 9.
If we substitute N = 10, d = 3 and k = 2 in the above theorem, we find that a natural number m = a_r * 10^r + … + a_1 * 10 + a_0 is divisible by 3 if and only if the sum of its digits a_r + … + a_1 + a_0 is divisible by 3.
As can be seen, there is nothing mysterious about this quite useful result, which we have shown to be true for every base greater than or equal to 2 and every number that has a power equal to that base – 1.
 Updated January 10th, 2017, 5:39 pm to add the following 
Still, you have to prove your assertion. Otherwise, it may turn out to be false for some really large numbers. In mathematics and logic, proofs must be that – proofs, not just assumptions based on empirical evidence. Experimentation should only be used to find candidates for new theorems or to find counterexamples to conjectures.Philosophy Explorer wrote: I challenge anyone to find an exception. Test it out on your computers and scientific calculators.
 Philosophy Explorer
 Posts: 2116
 Joined: May 25th, 2013, 8:41 pm
Re: The multiple of nine mystery
With (abc)n  (cab)n = r1, I let n be any natural number.
Further checking reveals divisibility by nine holds when n is any integer: positive, negative or zero.
PhilX
 Updated January 28th, 2017, 4:23 pm to add the following 
"Rule for Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. For large numbers this rule can be applied again to the result. In addition, the final iteration will result in a 9.
Examples
A.) 2,880: 2 + 8 + 8 + 0 = 18, 1 + 8 = 9, so 9 2,880.
B.) 3,564,213: 3+5+6+4+2+1+3=24, 2+4=6, so 9 does NOT divide 3,564,213.
Proof
The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.
For any integer x written as an· · · a3a2 a1a0 we will prove that if 9(a0 + a1+ a2+ a3 ... + an), then 9x and vice versa.
First, we can state that
x = a0 + a1×10 + a2×102 + a3×103... + an×10n
Next if we let s be the sum of its digits then
s = a0 + a1 + a2 + a3 + ... + an .
So
x  s = (a0  a0) + (a1 × 10  a1) + (a2×102  a2) + ... + (an×10n  an)
= a1(10  1) + a2(102  1) + ... + an(10n  1).
If we let bk = 10k  1, then bk = 9...9 (9 occurs k times) and bk =9(1…1) and we can rewrite the previous equation as
x  s = a1(b1)+ a2(b2)+ ... + an (bn)
It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk (which is xs) is also divisible by 9.
Since xs is divisible by 9, if x is divisible by 9, then so is s and vice versa."
Further checking reveals divisibility by nine holds when n is any integer: positive, negative or zero.
PhilX
 Updated January 28th, 2017, 4:23 pm to add the following 
Here is a proof for the divisibility by nine:Renee wrote:We were taught in Grade 7 in Hungary that you can check for divisibility by nine or by three by adding the digits, reductiowise, and if the last reduction results in 3 or 9, or in a number that is divisible by 3 or 9, then the original (nonreduced) number is divisible by 3 or nine.
I suspect it works for any positive integer power of three. (1, 3, 9, 27, 3^4, 3^5, ... 3^n where n is a positive integer.)
I don't know why this works. Nine is not mysterious. There is a proof somewhere. But don't ask "why" as if there were a divine decree about numbers.
Actually, I am wrong. The real reason this works is because Pope Poncificatius XIV in the eighth century Holy Roman Empire decreed it in his inaugural Easter sermon.
 Updated November 15th, 2016, 8:29 am to add the following 
LuckyR wrote:Does this hold true in base 6?That was a trick question. There is no digit "9" in a basesix number system.Dolphin42 wrote:Yes. Try it and see.
Plus, you are wrong. (Sorry.)
In base6,
nine=13
and an example of 9*4=36 3+6=9 does not work in base six;
13*4=100 and 1+0+0 does not equal 13.
"Rule for Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. For large numbers this rule can be applied again to the result. In addition, the final iteration will result in a 9.
Examples
A.) 2,880: 2 + 8 + 8 + 0 = 18, 1 + 8 = 9, so 9 2,880.
B.) 3,564,213: 3+5+6+4+2+1+3=24, 2+4=6, so 9 does NOT divide 3,564,213.
Proof
The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.
For any integer x written as an· · · a3a2 a1a0 we will prove that if 9(a0 + a1+ a2+ a3 ... + an), then 9x and vice versa.
First, we can state that
x = a0 + a1×10 + a2×102 + a3×103... + an×10n
Next if we let s be the sum of its digits then
s = a0 + a1 + a2 + a3 + ... + an .
So
x  s = (a0  a0) + (a1 × 10  a1) + (a2×102  a2) + ... + (an×10n  an)
= a1(10  1) + a2(102  1) + ... + an(10n  1).
If we let bk = 10k  1, then bk = 9...9 (9 occurs k times) and bk =9(1…1) and we can rewrite the previous equation as
x  s = a1(b1)+ a2(b2)+ ... + an (bn)
It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk (which is xs) is also divisible by 9.
Since xs is divisible by 9, if x is divisible by 9, then so is s and vice versa."