/sci/ - Science & Math

>>15102069
too complicated, wtf "mod" even is lol, speak in English bro

>>15102051
very simple bro

notice any number in the decimal system is composed of place values and digits that tell how many place values you have in each position. each place value is a power of 10 (for example, 123 means 1 * 10^2 + 2 * 10^1 + 3 * 10^0, this is because you compose a unit of higher value at a rate of 10 individual units, and this in turn is because you only have 0 to 9 digits to represent units).

if a number is divisible by 3 in the decimal system, that means each of the products of digit * respective power of 10 (place value in each position) will also all be divisible by 3, otherwise you're left with a non-integer remainder in the division.

the problem is, powers of 10 aren't divisible by 3, you will always be left with 1/3 (for example, 100/3 = 33 1/3, 10/3 = 3 1/3), so in any given division of a number n you have something like

a * 10^0 + b * 10^1 + c * 10^2 + ...

where a, b, c, etc. are natural numebrs called digits that go from 0 to 9

if you divide it by 3, you get

(a * 10^0 + b * 10^1 + c * 10^2 + ... ) / 3 = (a * 10^0) / 3 + (b * 10^1) / 3 + (c * 10^2) / 3 + ... =

= a * (1/3) + b * (3 + 1/3) + c * (33 + 1/3) + ... = a/3 + b/3 + 3b + c/3 + 33c + ... =

= 1/3 * (a + b + c + ... ) + 3b + 33c + ...

As you can see, this result of the division of the number by 3 will only be a natural number, and thus the number will be divisible by 3, if and only if the sum of the digits i.e. (a + b + c + ... ) is a multiple of 3 too