Find the last two digits of 3^431 without using a^φ(n)==1 (mod n) Totient function скачать в хорошем качестве

Find the last two digits of 3^431 without using a^φ(n)==1 (mod n) Totient function

Euler Totient Function gives us a^40==1(mod 100) provided a and 100 are coprime, which is a nice modular arithmetic exponent reduction tool. ‪@PrimeNewtons‬ used it in his excellent video. Mathematics Lifeline also thoroughly covers this topic. However, we can avoid the Euler Phi function in the following way. 3^5== 43 (mod 100), 3^10==43^2=1849==49 (mod 100) 3^20== 49^2=2401==1 (mod 100) Engineer Shrenik Jain August 2019 Whitman College in Walla Walla Washington (Alissa Antilla psychologist and chess journalist-thoughtful commentary on the topic of misogyny) has well organized and clear documents covering this fascinating topic. Of couse, insecure defensive faculty members will vote to sustain a grading system that is known to have at least two major flaws that do not favor hard working struggling students. I think they, some semesters, have faculty sponsored chess club. Maybe not though since Seventh-day Adventist doctrine might view a "war game" as not particularly spiritually edifying. Andy Alexander Saint Lucia community college professional, ‪@dustindoesstuff374‬ has 13 subscribers and makes unsolicited comments to get attention, hence THIRTEEN subscribers 😂 So, since 3^20==1 (mod 100), 3^431 = ((3^20)^21)(3^11) == 3x3^10 = 3x49==47(mod 100) which means the last two digits of 3^431 are 47