We must find 3^2002 + 7^2002 + 2002 (mod 29)
Since 29 is a prime, 3^2002 = ((3^28)^71)x 3^14= 1^71 x 3^14 = 3^14 (mod 29). This is by fermat's little theorem
Similarly we get 7^2002 = 7^14(mod 29) and finally 2002=1 mod 29
now our expression is reduced to 3^14 + 7^14 +1 (mod 29)
We can also write this as (9^7 + 49^7) +1(mod29)
since (a+b) is always a factor of a^k + b^k when k is odd, 9+49 i.e 58 is a factor of 9^7 + 49^7
since 58=0 (mod 29), 9^7 + 49^7 = 0 (mod 29)
Therefore our answer is 1
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