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Math Challenge II-A Number Theory

 
 
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Math Challenge II-A Number Theory
by Neo Liang - Saturday, 13 July 2019, 7:48 PM
 

Lecture 7, Problem 7.24

For problem 7.24, I got 13, but the answer said 8. It is 13 because 222=1 (mod 23). Then 22016=(222)91·2^7=2^7 (mod 23). 2^7=(2^5)*(2^2)=4*9=36=13 (mod 23).

 
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Re: Math Challenge II-A Number Theory
by Areteem Professor - Monday, 15 July 2019, 12:21 PM
 

Please note that $$2016 = 22\times 91 + 14,$$ so $$2^{2016} = \left(2^{22}\right)^{91}\cdot 2^{14}.$$ You can carry out the rest and verify what the answer actually is.

 
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