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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.


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Why I Hate Mathematicians
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