pack. To cut it, when arranged in its proper order, and transpose the two parts, is to add a constant amount to the face-value of every card. So much for addition.
Now how shall we multiply? Suppose we have the pack of ten in its proper order, and wish to multiply the face-value of the cards by 3. We deal out the cards one by one from first to last, into 3 piles, laying them face up upon the table. We first take up the pile the 10, or 0, falls upon, then the next pile, last the third. Putting each pile after the first at the back of that last taken.
We now find in place 1 card 3, or 3 times 1;
in place 2 card 6, or 3 times 2;
in place 3 card 9, or 3 times 3;
in place 4 card 2, congruent to 3 times 4;
in place 5 card 5, congruent to 3 times 5;
in place 6 card 8, congruent to 3 times 6;
in place 7 card 1, congruent to 3 times 7;
in place 8 card 4, congruent to 3 times 8;
in place 9 card 7, congruent to 3 times 9;
in place 10 card 0, congruent to 3 times 10.
Take this pack and multiply again by 3. Multiplying by 3 twice is multiplying by 9. But 9 ≡ − 1.
Accordingly we shall now find
in place 1 card − 1 or 9,
in place 2 card − 2 or 8,
in place 3 card − 3 or 7,
etc.
Multiply again by 3, and since 3 × 9 ≡ 7, we shall find
in place 1 card 7 × 1 ≡ 7,
in place 2 card 7 × 2 ≡ 4,
in place 3 card 7 × 3 ≡ 1,
in place 4 card 7 × 4 ≡ 8,
in place 5 card 7 × 5 ≡ 5,
in place 6 card 7 × 6 ≡ 2,
etc.
Take a pack of 11 cards. We shall now have,
11 ≡ 0
12 ≡ 1
23 ≡ 1
and, in short, to find what any card will be, having performed the necessary arithmetical operation, we subtract the number in the tens place from the number in the units place, and repay anything we borrow in the addition. Thus, suppose we deal into 5 piles, and take up the piles from left to right putting each one at the back of the pile that was at the left of it. We shall now have
in place 1, since 5 × 1 = 5, card 5;
in place 2, since 5 × 2 = 10, card 10;
in place 3, since 5 × 3 = 15 and 1 from 5 leaves 4, card 4;
in place 4, since 5 × 4 = 20 and 2 from 10 leaves 8, and repaying 1 borrowed we have 9, card 9;
in place 5, since 5 × 5 = 25 and 2 from 5 leaves 3, card 3;
in place 6, since 5 × 6 = 30 and 3 from 10 leaves 7, and repaying 1 we get 8, card 8;
in place 7, since 5 × 7 = 35 and 3 from 5 leaves 2, card 2;
in place 8, since 5 × 8 = 40 and 4 from 10 leaves 6, and repaying 1 we get 7, card 7;
in place 9, since 5 × 9 = 45 and 4 from 5 leaves 1, card 1;
in place 10, since 5 × 10 = 50 and 5 from 10 leaves 5, and repaying 1 we get 6, card 6;
in place 11, since 5 × 11 = 55 and 5 from 15 leaves 10, and repaying 1 we get 11, card 11 (≡ 0).
Suppose we now deal again into 9 piles. Now, the last card falls on the second pile. How are we to take up the piles? Answer: After the cards are exhausted, go on dealing in rotation upon the piles to the right of the last single card dealt no longer single cards but whole piles, always taking the extreme left hand one. Thus, in the present case, after the piles are all dealt out, put the left hand pile upon the pile to the right of the Jack, the last single card dealt; that is, put the pile headed by the 6 on that headed by the 4. Then, on the pile one further to the right, that headed by the 9, put the extreme left one headed by the Jack. Next, on the one headed by the 3 put the one headed by the 6, and so on until the piles are reduced to one. You will then find the proper order restored. Why? Because you have multiplied by 5 and by 9, that is, by 45, and 4 from 5 leaves 1, so that you have multiplied the cards in their proper order by 1, which leaves them in their proper order.
I now beg you, my dear Barbara, to take the full pack of 53 cards, and arrange them in their proper order, first the spades, second the diamonds, third the clubs, and fourth the hearts, each suit in its proper order,
1 2 3 4 5 6 7 8 9 X J Q K
with the Joker at the face. Deal them out into 12 piles and take up the piles according to the rule. Namely, denoting the Joker by O,
Next deal the cards out again into 31 piles, and take up the piles according to the rule. Namely,
This restores the original order because 12 × 31 = 372, and 53 into 372 goes 7 times and 1 over; so that
12 × 31 ≡ 1 (mod 53);
that is, the two dealings are equivalent to multiplying by 1; that is, they leave the cards in their original order.
You, Barbara, come from an ancient and a proud family. Conscious of being raised above the necessity of using ideas, you scorn them in your own exalted circle, while excusing them in common heads. Your cousins Baroco and Bocardo were always looked upon askance in the family, because they were suspected of harboring ideas,—a quite baseless suspicion, I am sure. But do you know that the unremitting study of years has tempted me to favor a belief subversive of your kindred’s supremacy, and of those principles of logic that are accepted upon all hands, I mean a belief that one secret of the art of reasoning is to think? In this matter of card-multiplication, instead of conceiving the dealing out into piles as one operation and the gathering in as another, I would prefer a general formula which shall describe both processes as one. At the outset, the cards being in no matter what order, we may conceive them as spread out into a row of 53 piles of 1 card each. If the cards are in their proper order, the last card is the Joker. In any case, you will permit me to call any pile that it may head the Ultima. The dealing out of the cards may be conceived to begin by our taking piles (single cards, at first) from the beginning of the row and putting them down in successive places following the ultima, until we reach the pile which we propose to make the final one, and which is destined to receive all the cards. When in this proceeding, we have reached the final pile, let us say that we have completed the first “round.” Thereupon we go back to the pile after the ultima as the next one upon which we will deposit a pile. We may complete a number of rounds each ending with placing a pile (a single card) on the final pile. We make as many as the number of cards in the pack will permit, and we will call these the rounds of the “first set.” It will be found useful, by the way, to note their number. Having completed them, we go on just as if we were beginning another; but when we have moved the ultima, let us say that we have completed the first round of the second set. Every round of the first set ends by placing a pile on the final pile. Let us call such a round “a round of the odd kind.” Every round of the second set ends by moving the ultima. Let us call such a round a round of the even kind. We make as many rounds of this kind as the whole number of places after the ultima enables us to complete. We call these the rounds of the second set. We then return to making rounds of the odd kind and make as many as the number of piles before the ultima enables us to make. So we go alternating sets of rounds of the odd and the even kind, until finally the ultima is placed upon the final