# Why I Hate Mathematicians

I was given a question:

`1234567891011121314151617181920 / 3`

`Is the remainder 0 or not 0?`

I did some classic long division to solve this question:

`1234567891011121314151617181920 / 3`

`= 41522630337040438050539060640`

`R = 0`

I thought I did it quickly, but apparently there is a much faster way. Add each digit in `1234567891011121314151617181920`

together:

`1+2+3+4+5+6+7+8+9 = 45`

`1+2+3+4+5+6+7+8+9 = 45`

`1+1+1+1+1+1+1+1+1+1 = 10`

`0+0+2 = 2`

`45+45+10+2 = 102`

Now add each digit in 102 together:

`1+0+2 = 3`

3 is divisible by 3, therefore `1234567891011121314151617181920`

is divisible by 3.

# Prove it

Expand 'abcd' where a, b, c, and d are digits 0 through 9:

`'abcd' = a*1000 + b*100 + c*10 + d`

1000 can also be expressed as 999+1, so:

`'abcd' = a*(999+1) + b*(99+1) + c*(9+1) + d`

999 can be expressed as 3*333, so:

`'abcd' = a*((3*333)+1) + b*((3*33)+1) + c*((3*3)+1) + d`

Multiply a, b, and c in:

`'abcd' = ((3*333)a+a) + ((3*33)b+b) + ((3*3)c+c) + d`

`'abcd' = (3*333*a+a) + (3*33*b+b) + (3*3*c+c) + d`

`'abcd' = (3*333a+a) + (3*33b+b) + (3*3c+c) + d`

Factor out 3, and reorganize:

`'abcd' = 3(333a+33b+3c) + (a+b+c+d)`

That's a bit fucked up, but it checks out. That still does not prove that this division cheat actually works in all cases.

## Now presenting: The Division Algorithm

`a / d = Q + R`

, so `a = d * Q + R`

Where:

`d`

is the **divisor**

`a`

is the **dividend**

`Q`

is the **quotient**

`R`

is the **remainder**

## The Modulo Operation

`a ≅ R (mod d)`

Returning to the previous section, we left off at:

`'abcd' = 3(333a+33b+3c) + (a+b+c+d)`

Since `a=dQ+R`

:

`a`

is **'abcd'**

`d`

is **3**

`Q`

is **(333a+33b+3c)**

`R`

is **(a+b+c+d)**

Then, considering the modulo operation, `a ≅ R (mod d)`

:

`'abcd' ≅ (a+b+c+d) (mod 3)`

# Conclusion

I hate mathematicians.

This division cheat works when your **divisor** is **3** or **9**.

## Support the Author

NotAwful is an opinionated student studying networking and information security. You can support their studies monthly via Patreon (USD), or directly via PayPal (CAD).