The multiple of nine mystery

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Mysterio448
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The multiple of nine mystery

Post by Mysterio448 » September 19th, 2016, 8:36 pm

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?

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Spiral Out
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Re: The multiple of nine mystery

Post by Spiral Out » November 12th, 2016, 9:09 am

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!
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Re: The multiple of nine mystery

Post by Dolphin42 » November 12th, 2016, 10:19 am

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 N-1 is equal to N-1. 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 N-1 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 N-1, the last digit has always fallen to 0. So the next addition of N-1, on this occasion, doesn't cause a carry from the last digit. It simply flips the last digit back up to N-1. i.e. it adds N-1 to the sum.

That sounded a lot clearer in my head than it looks on the page.

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LuckyR
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Re: The multiple of nine mystery

Post by LuckyR » November 14th, 2016, 8:37 pm

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?
Does this hold true in base 6?
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Re: The multiple of nine mystery

Post by Dolphin42 » November 14th, 2016, 8:57 pm

Yes. Try it and see.

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Re: The multiple of nine mystery

Post by Renee » November 15th, 2016, 8:20 am

We were taught in Grade 7 in Hungary that you can check for divisibility by nine or by three by adding the digits, reductio-wise, and if the last reduction results in 3 or 9, or in a number that is divisible by 3 or 9, then the original (non-reduced) 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?
Dolphin42 wrote:Yes. Try it and see.
That was a trick question. There is no digit "9" in a base-six number system.

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.

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Re: The multiple of nine mystery

Post by Dolphin42 » November 15th, 2016, 9:52 am

Renee:
That was a trick question. There is no digit "9" in a base-six number system.
See my post #3. In base 6, the relevant number is 5. In base N the relevant number is N-1.

As an example (in base 6):

5 * 4 = 32, 3 + 2 = 5

Try another one.

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Spiral Out
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Re: The multiple of nine mystery

Post by Spiral Out » November 15th, 2016, 6:32 pm

Dolphin42 wrote:Renee:
That was a trick question. There is no digit "9" in a base-six number system.
See my post #3. In base 6, the relevant number is 5. In base N the relevant number is N-1.

As an example (in base 6):

5 * 4 = 32, 3 + 2 = 5

Try another one.
5 * 5 = 25, 2 + 5 = 7
5 * 6 = 30, 3 + 0 = 3

Doesn't seem to work.
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Renee
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Re: The multiple of nine mystery

Post by Renee » November 15th, 2016, 11:01 pm

Dolphin42 wrote: As an example (in base 6):

5 * 4 = 32, 3 + 2 = 5

Try another one.
Spiral Out wrote:5 * 5 = 25, 2 + 5 = 7
5 * 6 = 30, 3 + 0 = 3

Doesn't seem to work.
Dolphin is actually right, Spiral.
5*5 in base six is 41. 4+1 = 5.
Ignorance is power.

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Re: The multiple of nine mystery

Post by Philosophy Explorer » November 29th, 2016, 6:43 pm

Here's another twist:

First I define anagram numbers (note the plural) by using a specific three-digit 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

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Re: The multiple of nine mystery

Post by Spiral Out » November 29th, 2016, 10:13 pm

Renee wrote:Dolphin is actually right, Spiral.
5*5 in base six is 41. 4+1 = 5.
Thanks for the clarification!
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Re: The multiple of nine mystery

Post by Philosophy Explorer » December 1st, 2016, 11:45 am

Philosophy Explorer wrote:Here's another twist:

First I define anagram numbers (note the plural) by using a specific three-digit 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
Let's take this to the next level.

As explained above, I've already said that abc - cab = r is (always) evenly divisible by nine. Now how about (abc)n - (cab)n = r1 where n is a natural number? I still maintain that r1 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 - (xn - yn) = remainder

It would be helpful to consider a more specific example:

(x - y)3 - (x3 - y3) =
-3x2y + 3xy2

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

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Re: The multiple of nine mystery

Post by ChanceIsChange » January 10th, 2017, 5:32 pm

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 --
Philosophy Explorer wrote: I challenge anyone to find an exception. Test it out on your computers and scientific calculators.
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.

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Re: The multiple of nine mystery

Post by Philosophy Explorer » January 25th, 2017, 4:24 pm

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 --
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, reductio-wise, and if the last reduction results in 3 or 9, or in a number that is divisible by 3 or 9, then the original (non-reduced) 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?
Dolphin42 wrote:Yes. Try it and see.
That was a trick question. There is no digit "9" in a base-six number system.

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.
Here is a proof for the divisibility by nine:

"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 9|x 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 x-s) is also divisible by 9.

Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa."

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