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Question Number 150327 by mathdanisur last updated on 11/Aug/21

1) 33x  ≡  48 (mod 654)  2) 5^(1000 000)   ≡  x (mod 41)  Find  x=?

$$\left.\mathrm{1}\right)\:\mathrm{33}\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{48}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{5}^{\mathrm{1000}\:\mathrm{000}} \:\:\equiv\:\:\boldsymbol{\mathrm{x}}\:\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\mathrm{Find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$

Answered by Rasheed.Sindhi last updated on 11/Aug/21

1) 33x  ≡  48 (mod 654)          33x  ≡  48+654×8 (mod 654)          33x  ≡  48+5232 (mod 654)          33x  ≡  5280 (mod 654)              x  ≡  160 (mod 654)

$$\left.\mathrm{1}\right)\:\mathrm{33}\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{48}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{33}\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{48}+\mathrm{654}×\mathrm{8}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{33}\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{48}+\mathrm{5232}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{33}\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{5280}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\:\:\equiv\:\:\mathrm{160}\:\left(\mathrm{mod}\:\mathrm{654}\right) \\ $$

Commented by mathdanisur last updated on 11/Aug/21

ThankYou Ser

$$\mathrm{ThankYou}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Aug/21

2) 5^(1000 000)   ≡  x (mod 41)         5^(10) ≡−1(mod 41)         (5^(10) )^2 ≡(−1)^2 =1(mod 41)         5^(20) ≡1(mod 41)         (5^(20) )^n ≡(1)^n (mod 41)           50^(20n) ≡1(mod 41)  ∀n∈N           5^(1000 000)   ≡  1 (mod 41)                  x≡1(mod 41)

$$\left.\mathrm{2}\right)\:\mathrm{5}^{\mathrm{1000}\:\mathrm{000}} \:\:\equiv\:\:\boldsymbol{\mathrm{x}}\:\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{10}} \equiv−\mathrm{1}\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{5}^{\mathrm{10}} \right)^{\mathrm{2}} \equiv\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{20}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{5}^{\mathrm{20}} \right)^{\mathrm{n}} \equiv\left(\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{50}^{\mathrm{20n}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{41}\right)\:\:\forall\mathrm{n}\in\mathbb{N} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{1000}\:\mathrm{000}} \:\:\equiv\:\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{41}\right) \\ $$

Commented by mathdanisur last updated on 11/Aug/21

ThankYou Ser

$$\mathrm{ThankYou}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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