Question Number 13584 by Tinkutara last updated on 21/May/17 $$\mathrm{Calculate}\:\mathrm{19}^{\mathrm{93}} \:\left(\mathrm{mod}\:\mathrm{81}\right). \\ $$ Commented by RasheedSindhi last updated on 22/May/17 $$\mathrm{Using} \\ $$$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \equiv\mathrm{1}+\mathrm{nx}\left(\mathrm{mod}\:\mathrm{y}\right) \\…
Question Number 13529 by Tinkutara last updated on 20/May/17 $$\mathrm{Let}\:{p}\:\mathrm{be}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:>\:\mathrm{3}.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:{p}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\mathrm{12}? \\ $$ Commented by prakash jain last updated on 20/May/17…
Question Number 13141 by tawa tawa last updated on 15/May/17 Commented by RasheedSindhi last updated on 15/May/17 $$\mathrm{Mr}\:\mathrm{tawa}\:\mathrm{tawa},\:\mathrm{you}\:\mathrm{are}\:\mathrm{requested} \\ $$$$\mathrm{to}\:\mathrm{use}\:\mathrm{bigger}\:\mathrm{font}.\mathrm{I}\:\mathrm{felt}\:\mathrm{much} \\ $$$$\mathrm{difficulity}\:\mathrm{to}\:\mathrm{read}\:\mathrm{this}\:\mathrm{question}. \\ $$$$\left(\mathrm{I}\:\mathrm{made}\:\mathrm{screen}\:\mathrm{shot}\:\mathrm{of}\:\mathrm{your}\:\mathrm{question}\:\right. \\…
Question Number 13121 by tawa tawa last updated on 14/May/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::\:\:\frac{\mathrm{2}}{\mathrm{15}}\:+\:\frac{\mathrm{2}}{\mathrm{35}}\:+\:\frac{\mathrm{2}}{\mathrm{63}}\:+\:\frac{\mathrm{2}}{\mathrm{99}}\:+\:…\:+\:\frac{\mathrm{2}}{\mathrm{9999}} \\ $$ Answered by nume1114 last updated on 15/May/17 $$\:\:\:\:\frac{\mathrm{2}}{\mathrm{15}}+\frac{\mathrm{2}}{\mathrm{35}}+\frac{\mathrm{2}}{\mathrm{63}}+\frac{\mathrm{2}}{\mathrm{99}}+…+\frac{\mathrm{2}}{\mathrm{9999}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}×\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}×\mathrm{7}}+\frac{\mathrm{2}}{\mathrm{7}×\mathrm{9}}+…+\frac{\mathrm{2}}{\mathrm{99}×\mathrm{101}} \\ $$$$=\underset{{k}=\mathrm{1}}…
Question Number 13102 by tawa tawa last updated on 14/May/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\::\:\:\mathrm{1}^{\mathrm{6}} \:+\:\mathrm{2}^{\mathrm{6}} \:+\:\mathrm{3}^{\mathrm{6}} \:+\:\mathrm{4}^{\mathrm{6}} \:+\:…\:+\:\mathrm{n}^{\mathrm{6}} \\ $$ Commented by tawa tawa last updated on 14/May/17…
Question Number 78567 by TawaTawa last updated on 18/Jan/20 Commented by mr W last updated on 18/Jan/20 Commented by mr W last updated on 18/Jan/20…
Question Number 425 by 123456 last updated on 25/Jan/15 $${x}^{\mathrm{2}} +{y}^{\mathrm{3}} ={z}^{\mathrm{4}} \:\mathrm{have}\:\mathrm{integer}\:\boldsymbol{\mathrm{solutions}}? \\ $$ Answered by prakash jain last updated on 02/Jan/15 $$\mathrm{Assuming}\:\mathrm{we}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:+\mathrm{ve} \\…
Question Number 131484 by bemath last updated on 05/Feb/21 $$\mathrm{Express}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{49}} {\sum}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}+\sqrt{\mathrm{n}^{\mathrm{2}} −\mathrm{1}}}}\:\mathrm{as}\:{a}+{b}\sqrt{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:{a}\:\mathrm{and}\:{b} \\ $$ Answered by benjo_mathlover last updated on 05/Feb/21 Answered…
Question Number 403 by 123456 last updated on 25/Jan/15 $$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2005}} }\:\mathrm{have}\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}? \\ $$ Answered by prakash jain last updated on 30/Dec/14 $$\mathrm{3}^{\mathrm{2003}} \:\mathrm{digits}\:\mathrm{will}\:\mathrm{be}\:\mathrm{present}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}. \\ $$$$\mathrm{10}^{{n}}…
Question Number 400 by 123456 last updated on 25/Jan/15 $$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{20}} }\:\mathrm{have}\:\mathrm{how}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{periodic}\:\mathrm{part}? \\ $$ Answered by prakash jain last updated on 29/Dec/14 $$\mathrm{3}^{\mathrm{20}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{or}\:\mathrm{5}.\:\mathrm{So} \\ $$$$\mathrm{periodicity}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{equation}…