kawaiipotato
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- HSC
- 2015
Re: HSC 2015 4U Marathon
![](https://latex.codecogs.com/png.latex?\bg_white $ Let a number be in the form: $ a_n a_{n-1} ... a_1 a_0 $ where $ a_n $ is the 'n'th digit $ )
![](https://latex.codecogs.com/png.latex?\bg_white a_n a_{n-1} ... a_1 a_0 = a_0 + 10*a_1 + 100*a_2 + ... + 10^{n-1}a_{n-1} + 10^{n}a_{n} )
![](https://latex.codecogs.com/png.latex?\bg_white = a_0 + (9+1)a_2 + ... + (9+1)^{n-1}a_{n-1} + (9+1)^{n}a_{n} )
![](https://latex.codecogs.com/png.latex?\bg_white = a_0 + (9+1)a_2 + ... + ( ^{n-1}C_0 9^{n-1} + ... + ^{n-1}C_{n-1} )a_{n-1} + ( ^{n}C_0 9^n + ... + ^{n}C_{n} )a_{n} )
![](https://latex.codecogs.com/png.latex?\bg_white = a_0 + a_1 + a_2 + ... + a_n + 9[ $ the rest $ ] $ since everything else will have a '9' in front $ )
![](https://latex.codecogs.com/png.latex?\bg_white $ Dividing by 3, $ )
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{a_0 + a_1 + a_2 + ... + a_n}{3} + 3[ $ the rest $ ] )
![](https://latex.codecogs.com/png.latex?\bg_white $ So for it to be divisible by 3, $ a_0 + a_1 + a_2 + ... + a_n = 3k $ where k is an integer $ )
![](https://latex.codecogs.com/png.latex?\bg_white \therefore a_0 + a_1 + a_2 + ... + a_n $ must be divisible by $ 3 )
Is this correct method? My notation and method seems weird
Is this correct method? My notation and method seems weird
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