- #1

Math100

- 773

- 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 ##.