Can someone explain to me why this pattern appears when consolidating numbers within multiplication tables to single digits

sequences-and-series

This is something I came across about a year ago while playing with numbers in my head, I shared it with a couple friends who couldn't explain why it happens either, and since forgot about it until recently.

Two examples of consolidating a number to a single digit:

27 becomes 2 + 7 = 9, so 27 consolidates to 9.

47 becomes 4 + 7 = 11, then 11 becomes 1 + 1 = 2, so 47 consolidates to 2.

Pretty arbitrary, but when we apply this to the different multiplication tables a pattern emerges for each of them, and the patterns display a very interesting symmetry/negative correlation with each other. Also note that once complete every sequence loops through itself again infinitely.

An example of a table with it's consolidations beneath:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24…
2, 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, 6…

Okay, that should be all the pretext needed.
I will now write the main sequences down with my observations beneath.

Multiples of 0:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0…

(Multiples of 0 remain the same number)

Multiples of 1:
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5…

(Multiples of 1 ascend through each digit)

Multiples of 2:
2, 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, 6, 8, 1…

(Multiples of 2 ascend through the even numbers, then the odd numbers)

Multiples of 3:
3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6…

(Multiples of 3 ascend through the single digit multiples of 3)

Multiples of 4:
4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, 7, 2…

(Multiples of 4 have two alternating sequences of descending numbers. This can also be seen as one ascending sequence, ascending through even numbers then odd numbers, always skipping one number that should logically be next)

Multiples of 5:
5, 1, 6, 2, 7, 3, 8, 4, 9, 5, 1, 6, 2, 7…

(Multiples of 5 have two alternating sequences of ascending numbers. This can also be seen as one descending sequence, descending through odd numbers then even numbers, always skipping one number that should logically be next)

Multiples of 6:
6, 3, 9, 6, 3, 9, 6, 3, 9, 6, 3, 9, 6, 3…

(Multiples of 6 descend through the single digit multiples of 3)

Multiples of 7:
7, 5, 3, 1, 8, 6, 4, 2, 9, 7, 5, 3, 1, 8…

(Multiples of 7 descend through the odd numbers, then the even numbers)

Multiples of 8:
8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4…

(Multiples of 8 descend through each digit)

Multiples of 9
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9…

(Multiples of 9 remain the same number)

This is the end of our pattern of sequences. The more observant of you will have spotted the negative correlations between sequences I mentioned earlier.

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

These sequences are a reflection of each other. Continuing our sequences with higher tables will give the exact same patterns repeating each 9 multiples.

eg.
9 with 18
10 with 17
11 with 16
12 with 15
13 with 14

The only anomaly is 0, based on it's nature, always gives 0.
9 will always give 9.
18 will always give 9.
27 will always give 9 etc.

These are the only multiples without a negative correlation to their counterpart, but a positive correlation.

So finally, can anyone explain to me why we find these patterns? I've found absolutely no mention of this on the internet when I looked, I think because of how arbitrary the question is, but there must be a solid reasoning as to why this happens this way. Thank you for your time!

(One final footnote for those interested, I tried this with the prime numbers to see if I could find a pattern but to no avail, here are some of them:)

2, 3, 5, 7, 2, 4, 8, 1, 5, 2, 4, 1, 5, 7, 2, 8, 5, 7, 4, 8…

Best Answer

$\color {green}{\text{ Definition :}}$

What you mean is to find the Digital Root of the numbers . We calculate digital roots by repeated summation of the digits of a number until we reach a single value .

The more mathematical way to look at digital roots is by defining them as the remainder of a number $\mod 9$. (in Base $10$)

For eg .

$$47=4+7=11=1+1=2$$ which is equivalent to

$$\color{blue}{47 \,\equiv \,2\mod 9}$$

$\color{red}{ \text { Patterns in the numbers:}}$

  • Multiples of 1:

    • It is clear that in each multiple you add $1$ , to get other . Hence the order increases from $1$ to $9$ and then reaches back to $1.$

  • Multiples of 2:

    • It is clear that in each multiple you add $2$ , to get other . Hence the order increases from $2$ to $8$ with a step of $2$ and then reaches to $1.$ and again continues with a step of $2$.

  • Multiples of 3:

    • It is clear that in each multiple you add $3$ , to get other . Hence the order increases from $3$ to $9$ with a step of $3$ and then reaches back to $3.$

  • Multiples of 4:

    • In Case of $4$ , you add $4$ or in terms of modular arithmetics , you add $-5 \quad (\, 4\, \equiv -5 \, \equiv \mod 9)$. Hence the order decreases from $4$ to $8$ to $3$ with a step of $-5$

  • Multiples of 5:

    • In Case of $5$ , you add $5$ or in terms of modular arithmetics , you add $-4 \quad (\, 5\, \equiv -4 \, \equiv \mod 9)$. Hence the order decreases from $5$ to $1$ to $6$ with a step of $-4$

    • Multiples of 6:

    • In Case of $6$ , you add $6$ or in terms of modular arithmetics , you add $-3 \quad (\, 6\, \equiv -3 \, \equiv \mod 9)$ Hence the order decreases from $6$ to $3$ to $9$ with a step of $-3$

  • Multiples of 7:

    • In Case of $7$ , you add $7$ or in terms of modular arithmetics , you add $-2 \quad (\, 7\, \equiv -2 \, \equiv \mod 9)$ . Hence the order decreases from $7$ to $5$ to $3$ with a step of $-2$

  • Multiples of 8:

    • In Case of $8$ , you add $8$ or in terms of modular arithmetics , you add $-1 \quad (\, 8\, \equiv -1 \, \equiv \mod 9)$ . Hence the order decreases from $8$ to $7$ to $6$ with a step of $-1$.

  • Multiples of 9:

    • For $9$ it is trivial to see that since all the terms are divisible by $ 9 $, the digital roots would be $9$ only.

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