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Question-81821




Question Number 81821 by ahmadshahhimat775@gmail.com last updated on 15/Feb/20
Answered by TANMAY PANACEA last updated on 15/Feb/20
1899=3×633              =3×3×211=3^2 ×(211)^1   number of devisor=(2+1)(1+1)=6  formula  N=a^p ×b^q ×c^r   nuber of devisor n=(p+1)(q+1)(r+1)  number of devisor  3,9,633,1899,211,1  pls check...
$$\mathrm{1899}=\mathrm{3}×\mathrm{633} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\mathrm{3}×\mathrm{211}=\mathrm{3}^{\mathrm{2}} ×\left(\mathrm{211}\right)^{\mathrm{1}} \\ $$$${number}\:{of}\:{devisor}=\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{6} \\ $$$${formula}\:\:{N}={a}^{{p}} ×{b}^{{q}} ×{c}^{{r}} \\ $$$${nuber}\:{of}\:{devisor}\:{n}=\left({p}+\mathrm{1}\right)\left({q}+\mathrm{1}\right)\left({r}+\mathrm{1}\right) \\ $$$${number}\:{of}\:{devisor} \\ $$$$\mathrm{3},\mathrm{9},\mathrm{633},\mathrm{1899},\mathrm{211},\mathrm{1} \\ $$$${pls}\:{check}… \\ $$

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