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Find out the mistake in Modulo Arithmetic

Please help me to find out an error.

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 \(7\).

Mark Hennings - 4 years, 2 months ago

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