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Question Number 202672 by BaliramKumar last updated on 31/Dec/23

$$\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{divisors}\mathrm{of}24\mathrm{is}60.$$$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{divisors}\mathrm{of}24\times 79.$$

Answered by a.lgnaoui last updated on 31/Dec/23

$$60$$

Commented by BaliramKumar last updated on 31/Dec/23

$$\mathrm{sum}\mathrm{of}\mathrm{divisors}\mathrm{of}1896=?[24\times 79=1896]$$

Commented by Panav last updated on 31/Dec/23

$$61$$

Answered by MM42 last updated on 31/Dec/23

$$d\mid 24\mathrm{\&}\mathrm{\Sigma }d=60$$$$d^{\prime }\mid 79\mathrm{\&}\mathrm{\Sigma }{d}^{\prime }=80$$$$\Rightarrow \mathrm{\Sigma }{d}^{″}=4800✓;d^{″}\mid 24\times 79$$

Commented by BaliramKumar last updated on 31/Dec/23

$$\mathrm{yes}$$

Commented by MM42 last updated on 31/Dec/23

$$letA=\{d;d\mid m\}\mathrm{\&}B=\{d^{\prime };d^{\prime }\mid n\}\mathrm{\&}C=\{d^{″};d^{″}\mid mn\};(m,n)=1$$$$\Rightarrow C=\{1,d_1,...,d_m,d_1^{\prime },...,d_n^{{}^{\prime }},d_1{d}_1^{{}^{\prime }},...,d_m{d}_n^{{}^{\prime }}\}$$$$\Rightarrow \mathrm{\Sigma }{d}^{{}^{″}}=\mathrm{\Sigma }d\times \mathrm{\Sigma }{d}^{\prime }$$

Commented by BaliramKumar last updated on 31/Dec/23

$$\mathrm{yes}\mathrm{sir}$$

Answered by BaliramKumar last updated on 31/Dec/23

$${\sigma }_1\left(24\right)=60$$$${\sigma }_1\left(79\right)=1+79=80$$$${\sigma }_1\left(24\times 79\right)=60\times 80=4800$$$${\sigma }_1\left(\mathrm{a}\right)=\mathrm{m}{\sigma }_1\left(\mathrm{b}\right)=\mathrm{n}$$$${\sigma }_1\left(\mathrm{ab}\right)=\mathrm{mn}\mathrm{Hcf}(\mathrm{a},\mathrm{b})=1\mathrm{\&}\mathrm{a}\ne \mathrm{b}$$$$$$

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