kingiyk:
Let's consider one example:
2. The Digital Root of any Trinity of Numbers(e.g. 111, 222, 333) is either 3, 6 , or 9.
111 » 1 + 1 + 1 = 3
222 » 2 + 2 + 2 = 6
333 » 3 + 3 + 3 = 9
This goes on for all eternity. Why is that?
This is just a mathematical fact. There's nothing remarkable about it. It's just a feature of numbers.
If we add the digits of a "trinity number", we get:
$$x+x+x=3x$$
That is, the sum of the digits is necessarily a number that is divisible by 3. Now, the sum of digits could be a single digit number or it could be a two digit number. Let's consider both cases:
Case 1. Suppose the sum of the digits is a 1 digit number.
Which one-digit numbers are divisible by 3? Answer: 3, 6, 9.
Case 2. Suppose instead that the sum of the digits is a 2 digit number, which we can write as "
ab", where
a is the first digit and
b is the second digit.
We know that:
$$10a + b = 3x.$$
where x is an integer. Now consider the sum of the digits of the 2 digit number:
a+b =(10a+b) - 9a
But we have already shown that 10a + b = 3x, so
$$a+b = (10a+b) - 9a = 3x - 9a = 3(x-3a)$$
This shows that the digital sum of the 2 digit number must, itself, be a multiple of 3. Now, either that sum is a single digit number, or it is another 2 digit number, but for the
new sum, we're back to exactly the same situation again. We can break it down into case 1 or case 2, rinse and repeat, to prove that the "digital sum" must be a multiple of 3, again, using the same arguments as above.
It follows that, because the only single digits that are multiples of 3 are 3, 6 and 9, the "digital sums" of
all "trinity numbers" must be either 3, 6 or 9.
What about repeated patterns? For instance, what about a number like 145145145, where the "pattern" is repeated 3 times?
Consider the "digital sum" of a number like that. It consists of three repeated "digital sums" of the same 3 digit number (in this example the number 145). Suppose the sum of the the digit number is y, then the big number sum of digits must be 3y, which is a multiple of 3, again.
The digital sum of the digits of a 9-digit number can only have a maximum of 2 digits. But we have already shown that every two-digit number that is divisible by 3 must have a digital sum that is also divisible by 3 (the proof is given above). Therefore, when we repeat the "digital sum" process, we must end up with a result of 3, 6 or 9, for the same reasons that were considered above.
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What's the significance of all this? Well, we've learned some interesting facts about numbers, maybe. What have we learned about God and the Holy Trinity? Nothing.
