Say you construct piles and the value of the top card used to generate pile has a value . Then pile has cards. The last pile as at most cards, but maybe less.
Choose 3 piles (not the last one), say their top card values are and so there coresponding stacks contain cards respectively. Choose any two top cards and add the values plus 10:
, (leaving the 3rd stack which has ).
Obviously you have cards in your hand (see whats happening?) So if you discard cards from your hand you are left with cards. Which is the face value of the top card of the third deck.
If you pick the lsat deck and it contains fewer cards than then obviously you have more than the correct number of cards in your hand and I don't think it works (did I miss something).