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If I can name any card and any position don't you need 52 ^ 2 decks? Or at least that many divided by 2 for counting off the top or bottom.

If you want to frame it that way, then each physical deck can handle 52 distinct queries. So you have 52^2 / 52 = 52 decks required.

No, because a single deck can answer for 52 queries. and 52 decks (offset 0-51) will be sufficient for any card and number combo

Oh, I see. For what it's worth, here is someone else performing the trick and it doesn't look like that's what's happening.

It still requires memorising the positions of 52^2 cards though.

No, because all decks have the same order (just rotated).