Question Number 145359 by imjagoll last updated on 04/Jul/21 $$\mathrm{How}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{will}\:\mathrm{there}\:\mathrm{be} \\ $$$$\mathrm{in}\:\mathrm{875}^{\mathrm{16}} \:? \\ $$ Answered by Olaf_Thorendsen last updated on 04/Jul/21 $$\mathrm{If}\:\mathrm{N}\:\mathrm{is}\:\mathrm{a}\:\mathrm{n}−\mathrm{digit}\:\mathrm{number}\:: \\ $$$$\mathrm{N}\:=\:{a}_{{n}−\mathrm{1}}…
Question Number 13744 by AH Soomro last updated on 23/May/17 $${Solve}\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}^{{x}} \equiv\mathrm{13}\left({mod}\:\mathrm{18}\right) \\ $$$${Pl}\:{give}\:{complete}\:{process}. \\ $$ Commented by prakash jain last updated on…
Question Number 13733 by prakash jain last updated on 22/May/17 $$\mathrm{Question}\:\mathrm{continuing}\:\mathrm{from} \\ $$$$\mathrm{mrW1}\:\mathrm{post}\:\mathrm{on}\:{p}^{\mathrm{2}} \:\mathrm{mod}\:\mathrm{n}\equiv\mathrm{1}. \\ $$$$\mathrm{Find}\:\mathrm{a}\:\mathrm{number}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{for}\:\mathrm{all}\:{m}<{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{HCF}\left({m},{n}\right)=\mathrm{1} \\ $$$${m}^{\mathrm{2}} \:\mathrm{mod}\:{n}\:=\mathrm{1} \\ $$$${e}.{g}.\:\mathrm{for}\:\mathrm{12}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{m} \\ $$$$\mathrm{are}\:\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11}.…
Question Number 13728 by prakash jain last updated on 22/May/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{n}!>\left(\frac{{n}}{\mathrm{3}}\right)^{{n}} \\ $$ Answered by mrW1 last updated on 23/May/17 $${Using}\:{mathematical}\:{induction}: \\ $$$${for}\:{n}=\mathrm{1}\:{we}\:{have}…
Question Number 13725 by prakash jain last updated on 22/May/17 $$\frac{\mathrm{1}}{\mathrm{7}}=.\overline {\mathrm{142857}} \\ $$$$\frac{\mathrm{1}}{\mathrm{7}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{recurring}\:\mathrm{decimal}\:\mathrm{of}\:\mathrm{period}\:\mathrm{6}. \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{20}} }? \\ $$ Commented by RasheedSoomro last…
Question Number 13724 by prakash jain last updated on 22/May/17 $$\mathrm{Expansion}\:\mathrm{of}\:\mathrm{1000}!\:\mathrm{has}\:\mathrm{249},\:\mathrm{0}'{s}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{non}−\mathrm{zero}\:\mathrm{digit}\:\mathrm{from} \\ $$$$\mathrm{right}. \\ $$$$\mathrm{1000}!=……{d}\mathrm{000}…\mathrm{00} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{d}? \\ $$ Commented by prakash jain…
Question Number 13606 by Tinkutara last updated on 21/May/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{19}^{\mathrm{93}} \:−\:\mathrm{13}^{\mathrm{99}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{162}. \\ $$ Answered by mrW1 last updated on 21/May/17 $$\mathrm{log}\:\left(\mathrm{19}^{\mathrm{93}} \right)=\mathrm{93}×\mathrm{log}\:\mathrm{19}\approx\mathrm{119}…
Question Number 13584 by Tinkutara last updated on 21/May/17 $$\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. \\…