Amusements in Mathematics eBook

Henry Dudeney
This eBook from the Gutenberg Project consists of approximately 597 pages of information about Amusements in Mathematics.

Amusements in Mathematics eBook

Henry Dudeney
This eBook from the Gutenberg Project consists of approximately 597 pages of information about Amusements in Mathematics.

383.—­THE “T” CARD PUZZLE.

If we remove the ace, the remaining cards may he divided into two groups (each adding up alike) in four ways; if we remove 3, there are three ways; if 5, there are four ways; if 7, there are three ways; and if we remove 9, there are four ways of making two equal groups.  There are thus eighteen different ways of grouping, and if we take any one of these and keep the odd card (that I have called “removed”) at the head of the column, then one set of numbers can be varied in order in twenty-four ways in the column and the other four twenty-four ways in the horizontal, or together they may be varied in 24 x 24 = 576 ways.  And as there are eighteen such cases, we multiply this number by 18 and get 10,368, the correct number of ways of placing the cards.  As this number includes the reflections, we must divide by 2, but we have also to remember that every horizontal row can change places with a vertical row, necessitating our multiplying by 2; so one operation cancels the other.

384.—­CARD TRIANGLES.

The following arrangements of the cards show (1) the smallest possible sum, 17; and (2) the largest possible, 23.

       1 7
      9 6 4 2
     4 8 3 6
    3 7 5 2 9 5 1 8

It will be seen that the two cards in the middle of any side may always be interchanged without affecting the conditions.  Thus there are eight ways of presenting every fundamental arrangement.  The number of fundamentals is eighteen, as follows:  two summing to 17, four summing to 19, six summing to 20, four summing to 21, and two summing to 23.  These eighteen fundamentals, multiplied by eight (for the reason stated above), give 144 as the total number of different ways of placing the cards.

385.—­“STRAND” PATIENCE.

The reader may find a solution quite easy in a little over 200 moves, but, surprising as it may at first appear, not more than 62 moves are required.  Here is the play:  By “4 C up” I mean a transfer of the 4 of clubs with all the cards that rest on it. 1 D on space, 2 S on space, 3 D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S on space, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5 D exchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H) on space (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1), 6 C up on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7 H on 8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C up on 5 D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 62 moves in all.  This is my record; perhaps the reader can beat it.

386.—­A TRICK WITH DICE.

All you have to do is to deduct 250 from the result given, and the three figures in the answer will be the three points thrown with the dice.  Thus, in the throw we gave, the number given would be 386; and when we deduct 250 we get 136, from which we know that the throws were 1, 3, and 6.

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Amusements in Mathematics from Project Gutenberg. Public domain.