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Find-last-two-digit-of-33-22-22-33-




Question Number 212033 by RojaTaniya last updated on 27/Sep/24
 Find last two digit of ∣33^(22) −22^(33) ∣
$$\:{Find}\:{last}\:{two}\:{digit}\:{of}\:\mid\mathrm{33}^{\mathrm{22}} −\mathrm{22}^{\mathrm{33}} \mid \\ $$
Answered by Frix last updated on 27/Sep/24
22^(33) −33^(22) =  199 502 557 353 381 471 378 605 391 983 038 967 797 408 063
$$\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} = \\ $$$$\mathrm{199}\:\mathrm{502}\:\mathrm{557}\:\mathrm{353}\:\mathrm{381}\:\mathrm{471}\:\mathrm{378}\:\mathrm{605}\:\mathrm{391}\:\mathrm{983}\:\mathrm{038}\:\mathrm{967}\:\mathrm{797}\:\mathrm{408}\:\mathrm{063} \\ $$
Commented by MATHEMATICSAM last updated on 27/Sep/24
Bruh
$$\mathrm{Bruh} \\ $$
Answered by A5T last updated on 27/Sep/24
22^(33) =2^(33) ×11^(33) =11^(22) (11^(11) )(8)^(11) >11^(22) ×3^(22) =33^(22)   ⇒∣33^(22) −22^(33) ∣=22^(33) −33^(22) ≡3(mod 4);  22^(33) −33^(22) ≡2−64≡13(mod 25)  ⇒22^(33) −33^(22) ≡63(mod 100)
$$\mathrm{22}^{\mathrm{33}} =\mathrm{2}^{\mathrm{33}} ×\mathrm{11}^{\mathrm{33}} =\mathrm{11}^{\mathrm{22}} \left(\mathrm{11}^{\mathrm{11}} \right)\left(\mathrm{8}\right)^{\mathrm{11}} >\mathrm{11}^{\mathrm{22}} ×\mathrm{3}^{\mathrm{22}} =\mathrm{33}^{\mathrm{22}} \\ $$$$\Rightarrow\mid\mathrm{33}^{\mathrm{22}} −\mathrm{22}^{\mathrm{33}} \mid=\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} \equiv\mathrm{3}\left({mod}\:\mathrm{4}\right); \\ $$$$\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} \equiv\mathrm{2}−\mathrm{64}\equiv\mathrm{13}\left({mod}\:\mathrm{25}\right) \\ $$$$\Rightarrow\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} \equiv\mathrm{63}\left({mod}\:\mathrm{100}\right) \\ $$

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