Your formulas are wrong. If $$c \equiv 7p + 11 \mod{27}$$ then by applying the modular arithmetic function $$c - 11 \equiv 7 p \mod{27}$$ and then rearrange the formula $$(c - 11) \times 7^{-1} \equiv p \mod{27}$$.

Therefore, the decryption function is: $$p \equiv (c - 11) \times 7^{-1} \mod{27}$$

For this we need to compute de multiplicative inverse of 7 modulo 27. Since they are coprime number, this inverse exists. And we can compute it using the Euler totient method like this:

$$7 ^{-1} \equiv 7 ^{\phi(27) -1} \mod{27}$$

$$\phi(27) = 27 \times (1 - \frac{1}{3}) = 18$$ then $$ 7^{-1} \equiv 7^{17} \equiv 4 \mod{27}$$

Finally the decode function is :

$$p \equiv (c - 11) \times 4$$

For your example:

$$(18 - 11) \times 4 \equiv 7 \times 4 \equiv 28 \equiv 1 \mod{27}$$

Your formulas are wrong. If $$c \equiv 7p + 11 \mod{27}$$ then by applying the modular arithmetic function $$c - 11 \equiv 7 p \mod{27}$$ and then $$(c - 11) \times 7^{-1} \equiv p \mod{27}$$.

Therefore, the decryption function is: $$p \equiv (c - 11) \times 7^{-1} \mod{27}$$

For this we need to compute de multiplicative inverse of 7 modulo 27. Since they are coprime number, this inverse exists. And we can compute it using the Euler totient method like this:

$$7 ^{-1} \equiv 7 ^{\phi(27) -1} \mod{27}$$

$$\phi(27) = 27 \times (1 - \frac{1}{3}) = 18$$ then $$ 7^{-1} \equiv 7^{17} \equiv 4 \mod{27}$$

Finally the decode function is :

$$p \equiv (c - 11) \times 4$$

For your example:

$$(18 - 11) \times 4 \equiv 7 \times 4 \equiv 28 \equiv 1 \mod{27}$$

Your formulas are wrong. If $$c \equiv 7p + 11 \mod{27}$$ then by applying the modular arithmetic function $$c - 1 \equiv 7 p \mod{27}$$ and then $$(c - 11) \times 7^{-1} \equiv p \mod{27}$$.

Therefore, the decryption function is: $$p \equiv (c - 11) \times 7^{-1} \mod{27}$$

For this we need to compute de multiplicative inverse of 7 modulo 27. Since they are coprime number, this inverse exists. And we can compute it using the Euler totient method like this:

$$7 ^{-1} \equiv 7 ^{\phi(27) -1} \mod{27}$$

$$\phi(27) = 27 \times (1 - \frac{1}{3}) = 18$$ then $$ 7^{-1} \equiv 7^{17} \equiv 4 \mod{27}$$

Finally the decode function is :

$$p \equiv (c - 11) \times 4$$

For your example:

$$(18 - 11) \times 4 \equiv 7 \times 4 \equiv 28 \equiv 1 \mod{27}$$

Your formulas are wrong. If $$c \equiv 7p + 11 \mod{27}$$ then by applying the modular arithmetic function $$c - 11 \equiv 7 p \mod{27}$$ and then rearrange the formula $$(c - 11) \times 7^{-1} \equiv p \mod{27}$$.

Therefore, the decryption function is: $$p \equiv (c - 11) \times 7^{-1} \mod{27}$$

For this we need to compute de multiplicative inverse of 7 modulo 27. Since they are coprime number, this inverse exists. And we can compute it using the Euler totient method like this:

$$7 ^{-1} \equiv 7 ^{\phi(27) -1} \mod{27}$$

$$\phi(27) = 27 \times (1 - \frac{1}{3}) = 18$$ then $$ 7^{-1} \equiv 7^{17} \equiv 4 \mod{27}$$

Finally the decode function is :

$$p \equiv (c - 11) \times 4$$

For your example:

$$(18 - 11) \times 4 \equiv 7 \times 4 \equiv 28 \equiv 1 \mod{27}$$