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}? \\ $$ Answered by mehdee7396 last updated on…
Question Number 220737 by Spillover last updated on 18/May/25 Answered by som(math1967) last updated on 18/May/25 $$\:\mathrm{0}.\overset{.} {\mathrm{6}}=\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{06}+\mathrm{0}.\mathrm{006}+… \\ $$$$=\frac{\mathrm{0}.\mathrm{6}}{\mathrm{1}−.\mathrm{1}}=\frac{\mathrm{6}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Commented by Spillover…
Question Number 220738 by Spillover last updated on 18/May/25 Answered by mr W last updated on 18/May/25 $${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} +…=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}={k} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}}…
Question Number 220159 by Nicholas666 last updated on 06/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}}…