Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 189700 by DAVONG last updated on 20/Mar/23

Commented by Rasheed.Sindhi last updated on 21/Mar/23

a=21

$${a}=\mathrm{21} \\ $$

Commented by mehdee42 last updated on 21/Mar/23

it^, s wrong .because 21^5 ≡^(100) 01⇒21^(2020) ≡^(100) 01⇒21^(2022) ≡^(100) 21^2 ≡^(100) 41

$${it}^{,} {s}\:{wrong}\:.{because} \\ $$$$\mathrm{21}^{\mathrm{5}} \overset{\mathrm{100}} {\equiv}\mathrm{01}\Rightarrow\mathrm{21}^{\mathrm{2020}} \overset{\mathrm{100}} {\equiv}\mathrm{01}\Rightarrow\mathrm{21}^{\mathrm{2022}} \overset{\mathrm{100}} {\equiv}\mathrm{21}^{\mathrm{2}} \overset{\mathrm{100}} {\equiv}\mathrm{41} \\ $$

Answered by mehdee42 last updated on 21/Mar/23

11^2 ≡^(100) 21⇒11^(10) ≡^(100) 21^5 ≡^(100) 01 ⇒11^(2020) ≡^(100) 01⇒11^(2022) ≡^(100) 21⇒a=11

$$\mathrm{11}^{\mathrm{2}} \overset{\mathrm{100}} {\equiv}\mathrm{21}\Rightarrow\mathrm{11}^{\mathrm{10}} \overset{\mathrm{100}} {\equiv}\mathrm{21}^{\mathrm{5}} \overset{\mathrm{100}} {\equiv}\mathrm{01} \\ $$$$\Rightarrow\mathrm{11}^{\mathrm{2020}} \overset{\mathrm{100}} {\equiv}\mathrm{01}\Rightarrow\mathrm{11}^{\mathrm{2022}} \overset{\mathrm{100}} {\equiv}\mathrm{21}\Rightarrow{a}=\mathrm{11} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/23

It′s a^(2021) , not a^(2022)

$${It}'{s}\:{a}^{\mathrm{2021}} ,\:{not}\:{a}^{\mathrm{2022}} \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/23

21^5 ≡1(mod 100) 21^(2020) ≡1(mod 100) 21^(2020) ×21≡21(mod 100) 21^(2021) ≡21(mod 100)

$$\mathrm{21}^{\mathrm{5}} \equiv\mathrm{1}\left({mod}\:\mathrm{100}\right) \\ $$$$\mathrm{21}^{\mathrm{2020}} \equiv\mathrm{1}\left({mod}\:\mathrm{100}\right) \\ $$$$\mathrm{21}^{\mathrm{2020}} ×\mathrm{21}\equiv\mathrm{21}\left({mod}\:\mathrm{100}\right) \\ $$$$\mathrm{21}^{\mathrm{2021}} \equiv\mathrm{21}\left({mod}\:\mathrm{100}\right) \\ $$

Answered by talminator2856792 last updated on 22/Mar/23

... X Y 1 × 11 = ... X Y 1 0 + ... Y 1 ^� ^� ^� ^� Y^� +1 ^� 1^� ⇒ 11^(...xyz1) = ...z1 11^(2021) = ...21 a = 11

$$\:...\:\mathrm{X}\:\mathrm{Y}\:\mathrm{1}\:×\:\mathrm{11}\:=\:\:...\:\mathrm{X}\:\mathrm{Y}\:\mathrm{1}\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:...\:\mathrm{Y}\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bar {\:}\bar {\:}\bar {\:}\bar {\:}\bar {\mathrm{Y}}+\mathrm{1}\bar {\:}\bar {\mathrm{1}} \\ $$$$\: \\ $$$$\:\:\Rightarrow\:\:\mathrm{11}^{...\mathrm{xyz1}} \:=\:...\mathrm{z1} \\ $$$$\: \\ $$$$\:\:\mathrm{11}^{\mathrm{2021}} \:=\:...\mathrm{21} \\ $$$$\:\:{a}\:=\:\mathrm{11} \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 22/Mar/23

11^(2021) ≠...21 11^(2021) =...11 : 11^(10) ≡1(mod 100) (11^(10) )^(202) ≡1^(202) (mod 100) (11^(2020) ×11≡1×11(mod 100) 11^(2021) ≡11(mod 11)

$$\mathrm{11}^{\mathrm{2021}} \neq...\mathrm{21} \\ $$$$\mathrm{11}^{\mathrm{2021}} =...\mathrm{11}\:: \\ $$$$\mathrm{11}^{\mathrm{10}} \equiv\mathrm{1}\left({mod}\:\mathrm{100}\right) \\ $$$$\left(\mathrm{11}^{\mathrm{10}} \right)^{\mathrm{202}} \equiv\mathrm{1}^{\mathrm{202}} \left({mod}\:\mathrm{100}\right) \\ $$$$\left(\mathrm{11}^{\mathrm{2020}} ×\mathrm{11}\equiv\mathrm{1}×\mathrm{11}\left({mod}\:\mathrm{100}\right)\right. \\ $$$$\mathrm{11}^{\mathrm{2021}} \equiv\mathrm{11}\left({mod}\:\mathrm{11}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com