- #1
Math100
- 768
- 219
- Homework Statement
- Establish the following divisibility criteria:
An integer is divisible by ## 3 ## if and only if the sum of its digits is divisible by ## 3 ##.
- Relevant Equations
- None.
Proof:
Let ## P(x)= \Sigma^{m}_{k=0} a_{k} x^{k} ## be a polynomial function.
Then ## N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0} ## for ## 0\leq a_{k}\leq 9 ##.
Since ## 10\equiv 1\pmod {3} ##, it follows that ## P(10)\equiv P(1)\pmod {3} ##.
Note that ## N\equiv (a_{m}+a_{m-1}+\dotsb +a_{1}+a_{0})\pmod {3} ##.
Thus ## 3\mid N\Leftrightarrow N\equiv 0\pmod {3}\Leftrightarrow P(10)\equiv 0\pmod {3}\Leftrightarrow P(1)\equiv 0\pmod {3} ##.
This means ## 3\mid P(1)\Leftrightarrow 3\mid (a_{m}+a_{m-1}+\dotsb +a_{2}+a_{1}+a_{0}) ##.
Therefore, an integer is divisible by ## 3 ## if and only if the sum of its digits is divisible by ## 3 ##.
Let ## P(x)= \Sigma^{m}_{k=0} a_{k} x^{k} ## be a polynomial function.
Then ## N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0} ## for ## 0\leq a_{k}\leq 9 ##.
Since ## 10\equiv 1\pmod {3} ##, it follows that ## P(10)\equiv P(1)\pmod {3} ##.
Note that ## N\equiv (a_{m}+a_{m-1}+\dotsb +a_{1}+a_{0})\pmod {3} ##.
Thus ## 3\mid N\Leftrightarrow N\equiv 0\pmod {3}\Leftrightarrow P(10)\equiv 0\pmod {3}\Leftrightarrow P(1)\equiv 0\pmod {3} ##.
This means ## 3\mid P(1)\Leftrightarrow 3\mid (a_{m}+a_{m-1}+\dotsb +a_{2}+a_{1}+a_{0}) ##.
Therefore, an integer is divisible by ## 3 ## if and only if the sum of its digits is divisible by ## 3 ##.