using Gauss' algorithm (for linear congruences) for A > B To solve $Bx \equiv A \pmod{m}$, use Gauss' algorithm.
The algorithm works perfectly when $A < B$. For example, to solve $6x \equiv 5 \pmod{11}$: $$x \equiv \frac{5}{6} \equiv \frac{5(2)}{6(2)} \equiv \frac{10}{12} \equiv \frac{10}{1}$$
so $x \equiv 10$
But when $A > B$, I run into problems. For example, trying to solve $7x \equiv 13 \pmod{100}$: $$x \equiv \frac{13}{7} \equiv \frac{13(15)}{7(15)} \equiv \frac{195}{105} \equiv \frac{95}{5} \equiv \frac{95(21)}{5(21)} \equiv \frac{1995}{105} \equiv \frac{95}{5}$$ and it continues to be $\frac{95}{5}$. Am I missing a step?
PS: I was applying the algorithm on random linear congruence problems I could find. The second example comes from http://www.johndcook.com/blog/2008/12/10/solving-linear-congruences/, which says the answer is $x \equiv 59$.
update
This answer answered my question. It explains that Gauss' algorithm works only on prime modulo.

Frank, you can use Gauss's Algorithm even if modulo is not prime. The only thing you need to take care is that multiplier should be co-prime to modulo.
Just keep multiplying denominator by a number so that denominator is near 100 till denominator become 1. However, the multiplier must be co-prime to 100.
$$\frac{13}{7} \pmod {100} \equiv \frac{13 × 29}{7 × 29} \pmod {100} \equiv \frac{-23}{3} \pmod {100}$$
$$ \equiv \frac {-23-100}{3} \pmod {100} \equiv \frac {-123}{3} \pmod {100} \equiv {-41} \pmod {100}$$
$$ \equiv {59} \pmod {100}$$
