Determine the order 7 modulo 60. Also find the inverse of 7 modulo 60 and solve the linear congruence
$$ 7x \equiv 59 \pmod {60} $$
Since $\gcd(7,60)=1$, Euler's Theorem tells us $7^{\phi(60)} \equiv 1 \pmod {60}$. And so the order is a factor of $\phi(60)=16$. These factors are 1, 2, 4, 8 and 16, and are the only ones we need to test.
The following calculations show $7^4 \equiv 2401 \equiv 1 \pmod {60}$.
| n | 7^n | 7^n mod 60 |
| 1 | 7 | 7 |
| 2 | 49 | 49 |
| 4 | 2401 | 1 |
The order of 7 modulo 60 is 4.
This means
$$ 7^4 \equiv 1 \pmod {60} $$
Factorising
$$ 7 \times (7^3) \equiv 1 \pmod {60} $$
So the inverse of 7 modulo 60 is $7^3 \equiv 343 \equiv 43 \pmod {60}$.
We can use the inverse to solve the linear congruence
$$ \begin{align} 7x & \equiv 59 \pmod {60} \\ \\ 7^{-1} \times 7x & \equiv 43 \times 59 \pmod {60} \\ \\ x & \equiv 2537 \pmod {60} \\ \\ & \equiv 17 \pmod {60} \end{align} $$
So the solution is $x \equiv 17 \pmod {60}$.