SOLUTION TO 349. STALEMATE. (3/3) Show more
|P|b|P| | |p| |q|
| |b| | |p|P|R|P|
| |P| |N|P| |P|Q|
| | |B| |K|B|N|R|
SOLUTION TO 349. STALEMATE. (2/3) Show more
9. R--KKt3 9. B--Kt6
10. P--QB4 10. P--B5
11. P--B3 11. P--K5
12. P--Q5 12. P--K6
And White is stalemated.
We give a diagram of the curious position arrived at. It will be seen that not one of White's pieces may be moved.
|r|n| | |k| |n|r|
| |p| | | | |p|p|
| | | |p| | | | |
|p| |p|P| | | | |
SOLUTION TO 349. STALEMATE. (1/3) Show more
Working independently, the same position was arrived at by Messrs. S. Loyd, E.N. Frankenstein, W.H. Thompson, and myself. So the following may be accepted as the best solution possible to this curious problem :--
1. P--Q4 1. P--K4
2. Q--Q3 2. Q--R5
3. Q--KKt3 3. B--Kt5 ch
4. Kt--Q2 4. P--QR4
5. P--R4 5. P--Q3
6. P--R3 6. B--K3
7. R--R3 7. P--KB4
8. Q--R2 8. P--B4
SOLUTION TO 304. BACHET'S SQUARE. (4/4) Show more
completed in 2 ways, making 1,152 ways in all.
SOLUTION TO 304. BACHET'S SQUARE. (3/4) Show more
assuming that the A, K, Q, J must be arranged in the form 1, and the D,
S, H, C in the form 2. He thus included reflections and half-turns, but not quarter-turns. They may obviously be interchanged. So that the correct answer is 2 × 576 = 1,152, counting reflections and reversals as different. Put in another manner, the pairs in the top row may be written in 16 × 9 × 4 × 1 = 576 different ways, and the square then
SOLUTION TO 304. BACHET'S SQUARE. (2/4) Show more
is one solution. But in each square we may put the letters in the top line in twenty-four different ways without altering the scheme of arrangement. Thus, in Diagram 4 the S's are similarly placed to the D's in 2, the H's to the S's, the C's to the H's, and the D's to the C's. It clearly follows that there must be 24×24 = 576 ways of combining the two primitive arrangements. But the error that Labosne fell into was that of
SOLUTION TO 304. BACHET'S SQUARE. (1/4) Show more
Let us use the letters A, K, Q, J, to denote ace, king, queen, jack; and D, S, H, C, to denote diamonds, spades, hearts, clubs. In Diagrams 1 and 2 we have the two available ways of arranging either group of letters so that no two similar letters shall be in line--though a quarter-turn of 1 will give us the arrangement in 2. If we superimpose or combine these two squares, we get the arrangement of Diagram 3, which
be done. The eminent French mathematician A. Labosne, in his modern edition of Bachet, gives the answer incorrectly. And yet the puzzle is really quite easy. Any arrangement produces seven more by turning the square round and reflecting it in a mirror. These are counted as different by Bachet.
Note "row of four cards," so that the only diagonals we have here to consider are the two long ones. (2/2)
304. BACHET'S SQUARE. One of the oldest card puzzles is by Claude Caspar Bachet de Méziriac,
first published, I believe, in the 1624 edition of his work. Rearrange the sixteen court cards (including the aces) in a square so that in no row of four cards, horizontal, vertical, or diagonal, shall be found two cards of the same suit or the same value. This in itself is easy enough,
but a point of the puzzle is to find in how many different ways this may (1/2)
SOLUTION TO 76. THE BARREL OF BEER. (2/2) Show more
20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).
SOLUTION TO 76. THE BARREL OF BEER. (1/2) Show more
Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these
76. THE BARREL OF BEER. A man bought an odd lot of wine in barrels and one barrel containing beer. These are shown in the illustration, marked with the number of gallons that each barrel contained. He sold a quantity of the wine to one man and twice the quantity to another, but kept the beer to himself. The puzzle is to point out which barrel contains beer. Can you say which one it is? Of course, the man sold the barrels just as he bought them, (1/2)
SOLUTION TO 272. THE NINE SCHOOLBOYS. (3/3) Show more
pairs of his 8). The reader may now like to try his hand at solving the two next cases of 21 boys on 15 days, and 33 boys on 24 days. It is,
perhaps, interesting to note that a school of 489 boys could thus walk out daily in one leap year, but it would take 731 girls (referred to in the solution to No. 269) to perform their particular feat by a daily walk in a year of 365 days.
SOLUTION TO 272. THE NINE SCHOOLBOYS. (2/3) Show more
under the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number of boys m. Then every boy will pair m-1 times, of which (m-1)/4 times he will be in the middle of a triplet and (m-1)/2 times on the outside. Thus, if we refer to the solution above, we find that every boy is in the middle twice (making 4 pairs) and four times on the outside (making the remaining 4
SOLUTION TO 272. THE NINE SCHOOLBOYS. (1/3) Show more
The boys can walk out as follows:--
1st Day. 2nd Day. 3rd Day.
A B C B F H F A G
D E F E I A I D B
G H I C G D H C E
4th Day. 5th Day. 6th Day.
A D H G B I D C A
B E G C F D E H B
F I C H A E I G F
Every boy will then have walked by the side of every other boy once and once only.
Dealing with the problem generally, 12n+9 boys may walk out in triplets
might be grouped on the first day as follows:--
A B C
D E F
G H I
Then A can never walk again side by side with B, or B with C, or D with E, and so on. But A can, of course, walk side by side with C. It is here not a question of being together in the same triplet, but of walking side by side in a triplet. Under these conditions they can walk out on six days; under the "Schoolgirls" conditions they can only walk on four days. (2/2)
272. THE NINE SCHOOLBOYS. This is a new and interesting companion puzzle to the "Fifteen Schoolgirls" (see solution of No. 269), and even in the simplest possible form in which I present it there are unquestionable difficulties. Nine schoolboys walk out in triplets on the six week days so that no boy ever walks _side by side_ with any other boy more than once. How would you arrange them?
If we represent them by the first nine letters of the alphabet, they (1/2)
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
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