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Question Number 211122 by hardmath last updated on 28/Aug/24

$$$$Is there a special rule for divisibility of four-digit numbers by 13?

Answered by mahdipoor last updated on 28/Aug/24

$$abcd\equiv 0\left(mod13\right)$$$$\Rightarrow 1000a+100b+10c+d\equiv 0$$$$\Rightarrow -a-4b-3c+d\equiv 0$$$$\Rightarrow a+4b+3c-d\equiv 0$$$$abcd\Rightarrow if13\mid (a+4b+3c-d)\Rightarrow 13\mid abcd$$$$forexample:$$$$1313\Rightarrow 13\mid 1+4\left(3\right)+3\left(1\right)-3=13$$$$6643\Rightarrow 13\mid 6+4\left(6\right)+3\left(4\right)-3=39$$$$andect...$$

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