Good day to you!

At first lets "transform/clarify" the statement a little-bit:

To find number of trailing zeroes in base-6, you have to find MINIMUM of "2" and "3" in factorization.

So you have N*M grid, you are on [0][0] corner and want to get to [?][M] /or/ [N][?], while maximizing the minimum of "2/3"

You know that you will travel through max 200 numbers < 2^10, meaning, that you can meet at most "200*9" two's (and offcourse less threes).

So now you are at [X][Y] and you can expand to [X+1][Y] /OR/ [X][Y+1]

You can also "REMEMBER" the number of "3" and number of "2" until now.

So first (more easier) lets have dynamic [N][M][1800][1200].

Now you can expand to [X+1][Y] / [Y][X+1], returning MAXIMUM {lets think about it recursively}, and whenever you reach "N" or "M", you return minimum(2s,3s).

In every node, lets have the number "factorized" to "2s" and "3s" — and with all expansions, add the number of "2s" / "3s".

Hope it is a little-bit clear (till now).

Anyway as you can see, 100*100*1800*1200 is too much, so what to do with this?

Well — a little trick — you know, that if you have "A+c 2s" and "A 3s" (or vice versa), you can expand (instead of [X][Y][A+c][A]) to ([X][Y][c][0]+c) [because you will always get result of "c" here]

YAY! This means, that you will NEVER have to store both [2s] and [3s] at one moment! {the second will be always 0}

So we can go with [N][M][1800][BOOL], where bool will tell us, whether we are actually storing 2s or 3s

The complexity will be N*M*1800*2 (since there is no other overhead).

I'm sorry (I'm not best at explaining) — hope you took at least a little "something" from it :P

Good Luck & Have nice day!