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You can quickly take note that gcd(110,20)=10. With this you can substitute 2 1 0 = x . Doing this gives the values of g c d ( x 1 1 − 1 , x 2 − 1 ) . Factoring both expressions we find: x 1 1 − 1 = ( x 1 0 + x 9 + . . . + x + 1 ) ( x − 1 ) x 2 − 1 = ( x + 1 ) ( x − 1 ) Notice that both expressions share a factor of x − 1 = 2 1 0 − 1 = 1 0 2 3