Question Number 223348 by cryptograph last updated on 21/Jul/25 $${Determine}\:{gcd}\left(\mathrm{13}{a}+\mathrm{19}{b},{ab}\right)\:{given}\:{that}\:{gcd}\left({a},\mathrm{19}\right)={gcd}\left({b},\mathrm{13}\right)=\mathrm{1} \\ $$ Commented by A5T last updated on 23/Jul/25 $$\mathrm{This}\:\mathrm{isn}'\mathrm{t}\:\mathrm{unique}.\:\mathrm{a}=\mathrm{1}\:\mathrm{and}\:\mathrm{b}=\mathrm{1}\Rightarrow\:\mathrm{gcd}\left(\mathrm{32},\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{a}=\mathrm{2}\:\mathrm{and}\:\mathrm{b}=\mathrm{2}\:\Rightarrow\:\mathrm{gcd}\left(\mathrm{64},\mathrm{4}\right)=\mathrm{4} \\ $$ Answered…
Question Number 223349 by cryptograph last updated on 21/Jul/25 $${let}\:{gcd}\left({n},{m}\right)=\mathrm{1}.\:{Determine}\:{gcd}\left(\mathrm{5}^{{m}} +\mathrm{7}^{{m}} ,\mathrm{5}^{{n}} +\mathrm{7}^{{n}} \right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223346 by cryptograph last updated on 21/Jul/25 $${proof}\:{gcd}\left(\mathrm{2}^{{m}} −\mathrm{1},\mathrm{2}^{{n}} −\mathrm{1}\right)=\mathrm{2}^{{gcd}\left({m},{n}\right)} −\mathrm{1} \\ $$ Answered by Raphael254 last updated on 22/Jul/25 $${if}\:{m}\:=\:{kn},\:{k}\:\in\:\mathbb{Z}^{\ast} : \\…
Question Number 223125 by BaliramKumar last updated on 15/Jul/25 Answered by mr W last updated on 16/Jul/25 $${A}\:{needs}\:\mathrm{85}{s}\:{for}\:{one}\:{round}, \\ $$$${B}\:{needs}\:\mathrm{45}{s}\:{for}\:{one}\:{round}. \\ $$$${case}\:\mathrm{1}:\:{they}\:{run}\:{in}\:{opposite}\:{directions} \\ $$$${say}\:{they}\:{meet}\:{after}\:{time}\:{t}\:{for}\:{the} \\…
Question Number 222736 by Lekhraj last updated on 06/Jul/25 Answered by Raphael254 last updated on 08/Sep/25 $$\begin{array}{|c|c|}{\boldsymbol{{To}}\:\boldsymbol{{a}}\:\boldsymbol{{number}}\:\boldsymbol{{be}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{33},}\\{\boldsymbol{{it}}\:\boldsymbol{{needs}}\:\boldsymbol{{to}}\:\boldsymbol{{be}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{3}\:\boldsymbol{{and}}\:\mathrm{11}}\\\hline\end{array} \\ $$$$ \\ $$$${Why}? \\ $$$$ \\ $$$${ab}\mid{n}\:\Rightarrow\:{a}\mid{n}\:{and}\:{b}\mid{n}…
Question Number 222352 by cryptograph last updated on 23/Jun/25 $${Prove}\:{that}\::\:\left({a}−{b}\right)\left({a}−{c}\right)\left({a}−{d}\right)\left({b}−{c}\right)\left({b}−{d}\right)\left({c}−{d}\right)\:{divisible}\:{by}\:\mathrm{12},\:{with}\:{a},{b},{c},{d}\:\in\mathbb{Z} \\ $$ Answered by vnm last updated on 23/Jun/25 $$\mathrm{Among}\:\mathrm{four}\:\mathrm{integers}\:\mathrm{there}\:\mathrm{will}\:\mathrm{always}\:\mathrm{be}\:\mathrm{two}\: \\ $$$$\mathrm{that}\:\mathrm{are}\:\mathrm{comparable}\:\mathrm{modulo}\:\mathrm{3}\: \\ $$$$\mathrm{and}\:\mathrm{two}\:\mathrm{pairs}\:\mathrm{or}\:\mathrm{three}\:\mathrm{that}\:\mathrm{are} \\…
Question Number 222249 by MrGaster last updated on 21/Jun/25 $$\mathrm{Prove}:\forall{n}\in\mathbb{Z}^{+} ,\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\ldots+{n}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}+\ldots+{n}\right)^{\mathrm{2}} \\ $$ Answered by MrGaster last updated on 21/Jun/25 Answered by…
Question Number 222097 by MathematicalUser2357 last updated on 17/Jun/25 Commented by mr W last updated on 17/Jun/25 $${wrong}! \\ $$$$\left({a}+{b}+{c}\right)^{{n}} =\underset{\underset{\mathrm{0}\leqslant{i},{j},{k}\leqslant{n}} {{i}+{j}+{k}={n}}} {\sum}\left(\frac{{n}!}{{i}!{j}!{k}!}{a}^{{i}} {b}^{{j}} {c}^{{k}}…
Question Number 221787 by efronzo1 last updated on 10/Jun/25 $$\:\:\sqrt{\frac{\mathrm{1}−\mathrm{4x}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}{\mathrm{2}}}\:=\:\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=?\: \\ $$ Answered by mr W last updated on 11/Jun/25 $$\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{2}{x}\right)^{\mathrm{2}}…
Question Number 221247 by fantastic last updated on 28/May/25 Commented by fantastic last updated on 28/May/25 $${what}\:{is}\:{the}\:{yellow}\:{chord}\:{length}? \\ $$ Answered by mehdee7396 last updated on…