What are the last ten digits of [math] 2015^{2014^{2013^{...^{3 ...

This is an old exam question from the introduction to abstract algebra course I took 7 years ago. I tried using the Chinese Remainder Theorem. I ...Read more

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let us write 2015 as 2015 = 2000 + 15 Lets say 2013^2014 = n So we have to find the last 4 digits of (2000 +15)^n Now lets consider the ...

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2013 is the first year since 1987 that has had four different numbers. 2013+1987=4000 4000 is a multiple of 1000 4000 / 1000Read more

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So the last 4 digits are 0625. There is considerably simpler approach. Firstly, both parts of a congruence and a modulus can be divided by their ...Read more

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Asides 2013, 2014 also has 3 prime factors (2x19x53). Even 2015 (5x13x31). Interestingly, the next time 3 consecutive years would have prime ...Read more

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Let's say a = 2015 and X=2014^{2013^{...^{3^{2}}}}. We basically want a^X \textrm{ mod } 10^{10}. Now, we can't calculate this even via fast modulo ...Read more

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Top movies of 2012,2013,2014,2015, and 2015 · 1. The Avengers · 2. The Dark Knight Rises · 3. The Amazing Spider-Man · 4. The Hobbit: An Unexpected Journey · 5.Read more

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let n=20152016−2017. If you reduce everything modulo 100 you'll get n≡152016−17(mod 100). To reduce 152016 , we calculate 152=125≡25(mod 100).Read more

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