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# Find out the mistake in Modulo Arithmetic

Case 1 : $$2222^{5555}=3^{5555}(mod7) = 243^{1111}(mod 7) = 5^{1111}(mod 7) = 5.5^{1110}(mod7)$$ $$= 5.625^{370}(mod 7) = 5.2^{370}(mod 7)= 5.2.2^{369}(mod 7)$$ $$= 10.8^{123}(mod 7) = 10.1^{123}(mod 7) = 10(mod 7) =3$$

Case 2: $$2222^{5555} = 3^{5555}(mod 7)= 3^{5}.3^{5550}(mod 7)= 3^{5}.729^{925}(mod 7)$$ $$= 3^{5}.1^{925}(mod7) = 3^{5}(mod 7) = 5$$

In the two cases we got different answer. Which is the wrong one? And where is the mistake? Please help me as I am very weak in this area. Thanks

Note by Kumar Ashutosh
4 years, 2 months ago

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Case 1 is wrong. Your mistake is to identify $$5\times5^{1110}$$ with $$5\times625^{370}$$ modulo$$7$$. Presumably you are trying to replace $$5$$ by $$625=5^4$$, but $$370 \neq 1100 \div 4$$.

Case 2 is right.

A much quicker way of getting there is to note that $$3^6 = 1$$ modulo $$7$$, and so $2222^{5555} \; = \; 3^{5555} \; = \; 3^5 \; = \; 243 \; = \; 5$ modulo $$7$$. We are using one case of a general result which states that $$x^{p-1} \,=\, 1$$ modulo $$p$$ for any prime $$p$$ and any integer $$x$$ which is coprime to $$p$$. This tells us that $$3^{5550} = 1$$ modulo $$7$$.

- 4 years, 2 months ago