<s>[INST] 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.
 [/INST] 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.
$$\frac{13}{7} \pmod {100} \equiv \frac{13 × 29}{7 × 29} \pmod {100} \equiv \frac{-23}{3} \pmod {100}$$
$$ \equiv \frac {-23-100}{3} \pmod {100} \equiv \frac {-123}{3} \pmod {100} \equiv {-41} \pmod {100}$$
$$ \equiv {59} \pmod {100}$$
</s>