The answer is 1.

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The number abc satisfies 0 ≡ 10

0 ≡ 10( 100a + 10b + c ) ≡ 100b + 10c + a ≡ bca (mod 27) such that bca is also divisible by 27. The simplification in step 4 holds true because 1000 = 999 + 1 ≡ 1 (mod 27) since 999 = 2737.