the four shall add up to a different sum. There will, of course, be four cards in the reduced pack that will not be used. These four may be any that you choose. It is not a difficult puzzle, but requires just a little thought. (2/2)
405. CARD MAGIC SQUARES. Take an ordinary pack of cards and throw out the twelve court cards. Now, with nine of the remainder (different suits are of no consequence) form the above magic square. It will be seen that the pips add up fifteen in every row in every column, and in each of the two long diagonals. The puzzle is with the remaining cards (without disturbing this arrangement) to form three more such magic squares, so that each of (1/2)
SOLUTION TO 356. QUEER CHESS. (2/2) Show more
impossible in the game of chess, because Black could not be given check by both rooks at the same time, nor could he have moved into check on his last move.
I believe the position was first published by the late S. Loyd.
SOLUTION TO 356. QUEER CHESS. (1/2) Show more
| | | | | | | | |
| | |R|k|R|N| | |
| | | | | | | | |
If you place the pieces as follows (where only a portion of the board is given, to save space), the Black king is in check, with no possible move open to him. The reader will now see why I avoided the term "checkmate,"
apart from the fact that there is no White king. The position is
would be a sufficient reason for my not doing so. (2/2)
356. QUEER CHESS. Can you place two White rooks and a White knight on the board so that the Black king (who must be on one of the four squares in the middle of the board) shall be in check with no possible move open to him? "In other words," the reader will say, "the king is to be shown checkmated."
Well, you can use the term if you wish, though I intentionally do not employ it myself. The mere fact that there is no White king on the board (1/2)
SOLUTION TO 309. THE FORTY-NINE COUNTERS. Show more
The counters may be arranged in this order:--
A1, B2, C3, D4, E5, F6, G7.
F4, G5, A6, B7, C1, D2, E3.
D7, E1, F2, G3, A4, B5, C6.
B3, C4, D5, E6, F7, G1, A2.
G6, A7, B1, C2, D3, E4, F5.
E2, F3, G4, A5, B6, C7, D1.
C5, D6, E7, F1, G2, A3, B4.
309. THE FORTY-NINE COUNTERS. Can you rearrange the above forty-nine counters in a square so that no letter, and also no number, shall be in line with a similar one,
vertically, horizontally, or diagonally? Here I, of course, mean in the lines parallel with the diagonals, in the chessboard sense.
SOLUTION TO 286. PAINTING THE DIE. Show more
The 1 can be marked on any one of six different sides. For every side occupied by 1 we have a selection of four sides for the 2. For every situation of the 2 we have two places for the 3. (The 6, 5, and 4 need not be considered, as their positions are determined by the 1, 2, and 3.) Therefore 6, 4, and 2 multiplied together make 48 different ways--the correct answer.
286. PAINTING THE DIE. In how many different ways may the numbers on a single die be marked,
with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4 must be on opposite sides? It is a simple enough question, and yet it will puzzle a good many people.
SOLUTION TO 255. THE LEVEL PUZZLE. (2/3) Show more
immediately below our starting point. That makes eight. If, however, we take the third route through the E on the diagonal, we then have the option of any one of the three V's, by means of each of which we may complete the word in four ways. We can therefore spell LEVEL in twelve ways through the diagonal E. Twelve added to eight gives twenty readings, all emanating from the L in the top left-hand corner; and as
SOLUTION TO 255. THE LEVEL PUZZLE. (1/3) Show more
Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight on to the V, from which letter the word may be completed in four ways, for there are four E's available through which we may reach an L. There are therefore four ways of reading through the right-hand E. It is also clear that there must be the same number of ways through the E that is
SOLUTION TO 341. THE FOUR FROGS. (7/7) Show more
last moved _from_." This is not, of course, the only way of placing the counters, but it is the simplest solution to carry in the mind.
There are several puzzles in this book that the reader will find lend themselves readily to this method.
SOLUTION TO 341. THE FOUR FROGS. (6/7) Show more
to the next vacant point (in either direction), where you deposit the counter. You proceed in the same way until all the counters are placed. Remember you always touch a vacant place and slide the counter from it to the next place, which must be also vacant. Now, by the "buttons and string" method of simplification we can transform the diagram into E. Then the solution becomes obvious. "Always move _to_ the point that you
SOLUTION TO 341. THE FOUR FROGS. (5/7) Show more
moves given above, and you will find that every little difficulty has disappeared.
In Diagram D I give another familiar puzzle that first appeared in a book published in Brussels in 1789, _Les Petites Aventures de Jerome Sharp_. Place seven counters on seven of the eight points in the following manner. You must always touch a point that is vacant with a counter, and then move it along a straight line leading from that point
SOLUTION TO 341. THE FOUR FROGS. (4/7) Show more
the diagram in the form shown in Diagram C, where the relationship between the buttons is precisely the same as in B. Any solution on C will be applicable to B, and to A. Place your white knights on 1 and 3 and your black knights on 6 and 8 in the C diagram, and the simplicity of the solution will be very evident. You have simply to move the knights round the circle in one direction or the other. Play over the
SOLUTION TO 341. THE FOUR FROGS. (3/7) Show more
moves are indicated by lines, to obviate the necessity of the reader's understanding the nature of the knight's move in chess. But it will at once be seen that the two problems are identical. The central square can, of course, be ignored, since no knight can ever enter it. Now,
regard the toadstools as buttons and the connecting lines as strings, as in Diagram B. Then by disentangling these strings we can clearly present
SOLUTION TO 341. THE FOUR FROGS. (2/7) Show more
This is the familiar old puzzle by Guarini, propounded in 1512, and I give it here in order to explain my "buttons and string" method of solving this class of moving-counter problem.
Diagram A shows the old way of presenting Guarini's puzzle, the point being to make the white knights change places with the black ones. In
"The Four Frogs" presentation of the idea the possible directions of the
SOLUTION TO 341. THE FOUR FROGS. (1/7) Show more
The fewest possible moves, counting every move separately, are sixteen. But the puzzle may be solved in seven plays, as follows, if any number of successive moves by one frog count as a single play. All the moves contained within a bracket are a single play; the numbers refer to the toadstools: (1--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4,
4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1).