Card Algebra - Futility Closet

From Henry Dudeney:

Take an ordinary pack of playing cards and regard all the court cards as tens. Now, look at the top card — say it is a seven — place it on the table face downwards and play more cards on top of it, counting up to twelve. Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile. Then look again at the top card of the pack — say it is a queen — then count 10, 11, 12 (three cards in all), and complete the second pile. Continue this, always counting up to twelve, and if at last you have not sufficient cards to complete a pile, put these apart. Now, if I am told how many piles have been made and how many unused cards remain over, I can at once tell you the sum of all the bottom cards in the piles. I simply multiply by 13 the number of piles less 4, and add the number of cards left over. Thus, if there were 6 piles and 5 cards over, then 13 times 2 (i.e. 6 less 4) added to 5 equals 31, the sum of the bottom cards. Why is this?

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Every pile must contain thirteen cards, less the value of the bottom card. Therefore, thirteen times the number of piles less the sum of the bottom cards, and plus the number of cards left over, must equal fifty-two, the number in the pack. Thus thirteen times the number of piles plus number of cards left over, less fifty-two, must equal sum of bottom cards. Or, which is the same thing, the number of piles less four, multiplied by thirteen, and plus the cards left over gives the answer as stated. The algebraically inclined reader can easily express this in terms of his familiar symbols. From Modern Puzzles and How to Solve Them, 1926.

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