Prove that (2222^5555)+(5555^2222) is divisible by 7
Proof:
we know
2222 mod 7 = 3
and 5555 mod 7 = 4 or -3
so (2222^5555)+(5555^2222) mod 7
= 3^5555 + (-3)^ 2222 mod 7
= 3^5555+ 3^2222 mod 7
= 3^2222(3^3333 + 1) mod 7
as 3^2222 is not divisible by 7 so
we need to show that
3^3333 + 1 mod 7 = 0
now as 7 is prime so as per format's little theorem
3^6 mod 7 = 1
3^3333 mod 7 = 3^(3333 mod 6) mod 7
= 3^ 3 mod 7 = 27 mod 7
so 3^3333 + 1 mod 7 = 28 mod 7 = 0
hence proved
4 comments:
this is very useful! :D but can you explain more on how you arrived at the 3^(3333 mod 6)mod 7 part? I mean, what gives you the 6 in mod 6? And how do we link this to the Fermat's theorem equation? Thank you and please try to reply by todayyy! :D THANKSALOT!
as 7 is co prime to 3
so 3^6 mod 7 = 1
so 3^x = 3^(x mod 6)
I realized that there is a simpler proof
2222= 3 mod 7
5555 = 4 mod 7
2222^5555 + 5555^2222 = (3^5)^1111 + (4^2)^ 1111
as it is a^n + b^n and n is odd so it is divisible by
3^5 + 4^ 2 = 259
hence by 7 as 7 is a factor of 259
This one is easier..thanks @kpt..
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