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You want to find an integer n such that 5n leaves a remainder 1 when divided by 27. Well, 5 times any integer ends in 0 or 5. So just find a multiple of 27 that ends in 4 or 9, and add 1 to it. 27 × 2 = 54; 54 + 1 = 55 = 5 × 11. So the multiplicative inverse of 5 in $\displaystyle \mathbb{Z}_{27}$ is 11.