Question Number 122896 by fajri last updated on 20/Nov/20 $${Prove}\:: \\ $$$$\left({n},\mathrm{7}\right)\:=\:\mathrm{1}\:\rightarrow\:\mathrm{7}\mid\left({n}^{\mathrm{6}} \:−\:\mathrm{1}\right) \\ $$ Answered by Snail last updated on 20/Nov/20 $${Use}\:\:{Euelers}\:{theorem}\:{or}\:{fermats}\:{theorem}\:{to}\:{do}\:{so} \\ $$…
Question Number 188196 by cortano12 last updated on 26/Feb/23 $$\left(\mathrm{1}\right)\mathrm{solve}\:\mathrm{Diopthantine}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\mathrm{754x}+\mathrm{221y}=\mathrm{13} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{abcd}\: \\ $$$$\:\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{4}×\left(\mathrm{abcd}\right)=\mathrm{dcba} \\ $$ Commented by ARUNG_Brandon_MBU last updated on 26/Feb/23…
Question Number 122637 by sina1377 last updated on 18/Nov/20 $${can}\:{anybody}\:{do}\:{me}\:{a}\:{faver}\:{and}\:{tell}\:{me}\:{what}\:\:{the}\:\:{direct}\:{proof}\:{of}\:\:{mobius}\:{inversion}\:{formula}\:{is}?\:\left({proof}\:{without}\:{using}\:\:{Dirichlet}\:{convolutions}\right)? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121834 by bemath last updated on 12/Nov/20 Answered by liberty last updated on 12/Nov/20 $$\mathrm{we}\:\mathrm{have}\:\mathrm{gcd}\left(\mathrm{93},\mathrm{27}\right)\:=\:\mathrm{3}\:\mathrm{and}\:\mathrm{6}\:\mathrm{divides}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{therefore}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{solutions}.\: \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{93}\:=\:\mathrm{3}×\mathrm{27}+\mathrm{12}\:\wedge\:\mathrm{27}\:=\:\mathrm{2}×\mathrm{12}+\mathrm{3} \\ $$$$\:\mathrm{12}\:=\:\mathrm{4}×\mathrm{3}\:+\:\mathrm{0}\: \\ $$$$\:\mathrm{3}\:=\:\mathrm{1}×\mathrm{27}−\mathrm{2}×\mathrm{12}\:…
Question Number 187300 by anurup last updated on 15/Feb/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\left({p}+{q}\right)\left({q}+{r}\right)\left({r}+{p}\right)=\mathrm{8}{pqr}+\mathrm{2} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187296 by anurup last updated on 15/Feb/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{4},\:\mathrm{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{k}!} \:\mathrm{is}\:\mathrm{never}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{cube}. \\ $$ Commented by anurup last updated on 15/Feb/23 $$\mathrm{Please}\:\mathrm{solve}\:\mathrm{this}. \\…
Question Number 121694 by bemath last updated on 11/Nov/20 $${Evaluate}\:\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{4}{k}+\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{2}{k}−\mathrm{1}}\:+\:\sqrt{\mathrm{2}{k}+\mathrm{1}}}\:? \\ $$ Answered by liberty last updated on 11/Nov/20 $$\:\mathrm{let}\:\sqrt{\mathrm{2k}−\mathrm{1}}\:=\:\mathrm{a}\:\mathrm{and}\:\sqrt{\mathrm{2k}+\mathrm{1}}\:=\:\mathrm{b}\: \\ $$$$\Rightarrow\:\mathrm{a}^{\mathrm{2}}…
Question Number 187213 by anurup last updated on 14/Feb/23 $$\mathrm{If}\:{a},{b}\:\mathrm{are}\:\mathrm{co}-\mathrm{prime}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{satisfying} \\ $$$${a}^{\mathrm{2}} \:+\:{b}\:=\:\left({a}−{b}\right)^{\mathrm{3}} \:\mathrm{and}\:{b}+\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{find}\:\mathrm{all} \\ $$$$\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\left({a},\:{b}\right). \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this}. \\ $$$$ \\ $$ Terms of Service…
Question Number 55701 by mathaddicted last updated on 02/Mar/19 $${Is}\:{true}\:{or}\:{not}\:{that}\:\mathrm{4181}\:{is}\:{the}\:{only}\:{one} \\ $$$${Fibonacci}'{s}\:{number}\:{with}\:{no}\:{prime}\:{factor} \\ $$$${which}\:{is}\:{also}\:{a}\:{Fibonacci}'{s}\:{number}? \\ $$ Commented by MJS last updated on 03/Mar/19 $$\mathrm{1346269}=\mathrm{557}×\mathrm{2417} \\…
Question Number 55631 by Tawa1 last updated on 28/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19 $$\left(\mathrm{1}−{x}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}×\mathrm{3}}{\mathrm{2}×\mathrm{4}}{x}^{\mathrm{2}} +\frac{\mathrm{1}×\mathrm{3}×\mathrm{5}}{\mathrm{2}×\mathrm{4}×\mathrm{6}}{x}^{\mathrm{3}} +… \\ $$$$\boldsymbol{{L}}{HS} \\ $$$$\mathrm{1}+\frac{\mathrm{4}}{\mathrm{10}}+\frac{\mathrm{4}×\mathrm{12}}{\mathrm{10}×\mathrm{20}}+\frac{\mathrm{4}×\mathrm{12}×\mathrm{20}}{\mathrm{10}×\mathrm{20}×\mathrm{30}}+……