Finding GCD is easy isn't it?
Number Theory
Level
3
Compute -
GCD( 2002 + 2 , 2002^2 + 2 , 2002^3 + 2 , .....)
The answer is 6.
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1 solution
Let g denote the desired greatest common divisor.
Note that 2002^2 +2 = 2002(2000+2)+2 = 2000(2002+2)+6. By the Euclidean algorithm, we have ,
gcd(2002 + 2, 2002^2 + 2) = gcd(2004, 6) = 6.
Hence g | gcd(2002 + 2, 2002^2 + 2) = 6.
On the other hand, every number in the sequence 2002 + 2, 2002^2 + 2, . . . is divisible by 2.
Furthermore, since 2002 = 2001 + 1 = 667 * 3 + 1, for all positive integers k, 2002^k = 3a(subscript k) + 1 for some integer a(subscript k) .
Thus 2002^k + 2 is divisible by 3
Because 2 and 3 are relatively prime, every number in the sequence is divisible by 6.
Therefore, g = 6.