One way to find a solution is to use the grid or virtual quantum computer in the following way:

Find the cell value at (e,1) where e is the remainder for c.

You are looking for a[t] = na

Remember

At that value, d[t] = na+x

Also

At that value, x[t] = x, the x value in the cell is equal to the x value at (e,n)

REMEMBER, the value of x at na in (e,1) is the SAME as x at (e,n)

REMEMBER, take d from all values of d[t] at (e,1) and there is a known patter of (n-1) as factor in these values of d[t]-d that is different (increasingly) from the pattern of factors of n in a[t]. It is THIS that gives the offset that is used to solve the problem and thus get the cell at (e,1) to do all the work for you.

This isn't exactly a "known pattern" but it's a (very vague) set of steps with which to use the VQC to factor arbitrarily-large numbers.