Remainders... :/
Number Theory
Level
3
2005^2002 + 2002^2005 divided by 200, Find Remainder.
The answer is 57.
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answer: 57
i found an interesting fact that:
(a+b) (mod c) = a (mod c) + b(mod c)
and:
a^b (mod c) = {[(a (mod c)]}^b
then,
(2005^2002 + 2002^2005) (mod 200)
= (2005^2002) (mod 200) + (2002^2005) (mod 200)
First...
(2005^2002) (mod 200)
----> (2005)^(40*50 + 2) (mod 200)
By Euler's Formula,
2005^(µ 200) ≡ 1 (mod 200)
2005^40 ≡ 1 (mod 200)
----> (2005)^(40*50 + 2) (mod 200)
----> (2005^40)^50 * (2005)^2 (mod 200)
----> 1^50 * 2005^2 (mod 200)
-----> (2005)/200 = 10 r. 5
-----> [(2005)^2]/200 = 20,100 r. 25
(2005^2002) (mod 200) = 25
2nd:
(2002^2005) (mod 200)
-----> (2002)^(40*50 + 5) (mod 200)
----> [(2002)^40]^50 * (2002)^5 (mod 200)
----> 1^50 * (2002)^5 (mod 200)
----> (2002)^5 (mod 200)
----> [2002 (mod 200)]^5
----> 2^5
----> 32
(2002^2005) (mod 200) = 32
From
(a+b) (mod c) = a (mod c) + b(mod c),
we can say now that:
(2005^2002 + 2002^2005) (mod 200)
= (2005^2002) (mod 200) + (2002^2005) (mod 200)
= 25 + 32
= 57