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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?

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

Answered by mahdipoor last updated on 28/Aug/24

abcd ≡ 0     (mod 13)  ⇒1000a+100b+10c+d ≡ 0   ⇒ −a−4b−3c+d ≡ 0   ⇒ a+4b+3c−d ≡ 0   abcd ⇒ if 13∣(a+4b+3c−d)  ⇒ 13∣abcd  for example :  1313 ⇒ 13∣1+4(3)+3(1)−3=13   6643 ⇒ 13∣6+4(6)+3(4)−3=39  and ect ...

$${abcd}\:\equiv\:\mathrm{0}\:\:\:\:\:\left({mod}\:\mathrm{13}\right) \\ $$$$\Rightarrow\mathrm{1000}{a}+\mathrm{100}{b}+\mathrm{10}{c}+{d}\:\equiv\:\mathrm{0}\: \\ $$$$\Rightarrow\:−{a}−\mathrm{4}{b}−\mathrm{3}{c}+{d}\:\equiv\:\mathrm{0}\: \\ $$$$\Rightarrow\:{a}+\mathrm{4}{b}+\mathrm{3}{c}−{d}\:\equiv\:\mathrm{0}\: \\ $$$${abcd}\:\Rightarrow\:{if}\:\mathrm{13}\mid\left({a}+\mathrm{4}{b}+\mathrm{3}{c}−{d}\right)\:\:\Rightarrow\:\mathrm{13}\mid{abcd} \\ $$$${for}\:{example}\:: \\ $$$$\mathrm{1313}\:\Rightarrow\:\mathrm{13}\mid\mathrm{1}+\mathrm{4}\left(\mathrm{3}\right)+\mathrm{3}\left(\mathrm{1}\right)−\mathrm{3}=\mathrm{13}\: \\ $$$$\mathrm{6643}\:\Rightarrow\:\mathrm{13}\mid\mathrm{6}+\mathrm{4}\left(\mathrm{6}\right)+\mathrm{3}\left(\mathrm{4}\right)−\mathrm{3}=\mathrm{39} \\ $$$${and}\:{ect}\:... \\ $$

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