system
Given the following post, respond as if you're commenting on Reddit:
human
ELI5: Why is the difference between the sum of a whole number's places and the number itself is ALWAYS a direct multiple of 9?

For example let's assume a number 142.
So 1+4+2=7

142-7=135,  which is a multiple of 9.
gpt
In the decimal system (which is the one we use), the places of a number are how it's written as a sum of powers of ten. For example 142 = 1×10^2 + 4×10^1 + 2×10^0

So if we apply your operation to a number that's written "abc" (**a** is the first digit, **b** is the second digit, **c** is the last digit), a little algebra can show us how it works:

(**a**×100 + **b**×10 + **c**) - (**a**+**b**+**c**) 

= **a**×100-**a** + **b**×10-**b** + **c**-**c**

= 99×**a** + 9×**b**

and now we divide by 9:

(99×**a** + 9×**b**) / 9 = 11**a**+**b**.

Since a and b are whole numbers, the final result will always be a whole number, so it's divisible by 9. 


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ELIhatealgebra:

Let's take a big number, for example 7342.

It's actually made of a group of 1000 sevens, a group of 100 threes, a group of 10 fours, and a single solitary two. That's what all big numbers are, the digits of a number just tell you exactly what groups of 10, 100, 1000 and so on it is made of! It's what we call the "decimal system", or "base ten". 

If you remove one instance of each of the digits, well you're left with 999 sevens, 99 threes, and 9 fours! (You remove the single solitary two, so there are no lonely digits left). So you're left with a number that is just a collection of groups of nine, ninety-nine, nine-hundred-and-ninety nine digits! These groups are all multiples of nine! So the number itself is, too a multiple of nine! (because if you divide each of the groups it contains by nine, all groups make "good" divisions with no leftover! 99/9 = 11, 999/9 =111 and so on, they'll always divide neatly).