Mike, I think Susan Jo is probably thinking of five steps in the traditional algorithm for 62x36 as follows: 6x2, 6x60 30x2, 30x60 then adding the sums
It's like the five steps used in multiplying binomials (four multiplications and then adding them all up): (a+b)(c+d) = ac + ad + bc + bd (60+2)(30+6)= 60(30) + 60(6)+ 2(30) + 2(6) = 2232
But the traditional algorithm is more efficient than the line above in the way that it records those 5 steps. It makes the final addition easier by combining two of the products at a time, taking advantage of place value: 6x2=12 and 6x60= 360 is recorded in a vertical format as 372 and 30x2=60 and 30x60=1800 is recorded as 1860 then you only have to add those two numbers at the end.
So we've avoided writing a lot of zeros and we only have to add two numbers at the end, but at what price? Unfortunately, the price of this efficiency is that many students don't get that they are really multiplying 30x60 not 3x6, and that the answer for that part of the problem is really 1800, not 18. And many students don't know why the heck they have to put a zero at the right end of the second line, which could be considered a sixth step in itself.
I guess Susan Jo's method in her example could also be considered 5 steps, if you count 62x30 as two steps, though she might be doing this in her head and writing down the answer all at once.
Anyway, I don't think counting up steps is a perfect way to measure efficiency of an algorithm, because some steps take longer than others, either in thinking or in recording. The traditional algorithms are certainly efficient in recording, but by avoiding writing zeros that show place value they put up a barrier to understanding.