I have to find the last two digits of 222227^2010. So I know that this gets reduced to 27^2010 mod 100 But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong?
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Originally Posted by HeadOnAPike I have to find the last two digits of 222227^2010. So I know that this gets reduced to 27^2010 mod 100 But I don't know how to solve 27^x = 1(mod 100) to get this process going. Or am I going about it wrong? and Now . Observe Thus Therefore the last two digits of are .
Originally Posted by chiph588@ and This is the part I don't get. What does the phi mean, and what does the (27,100) mean? How did you get that 27^40 = 1 so simply?
is the "totient function", the number of integers less than 100 and coprime to 100. Chiph588 then uses "Euler's theorem": For any a coprime to n, .
What's a quick way to figure out the totient function?
Therefore , Originally Posted by HeadOnAPike What's a quick way to figure out the totient function? You may show that is a bijection . Then , obtaining the product :
Originally Posted by simplependulum What's going on here?
Hello, HeadOnAPike! I used a very primitive approach . . . Find the last two digits of: . I know that this gets reduced to: . We have: . I found that: . . Hence: . We have: . Hence: . . . . . . . . . . . . . . . . . Therefore, ends in 49.
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