outward journey in the short time of ten minutes, though it took him an hour to get back to the starting point at Slocomb, with the wind dead against him. Now, how long would the ten miles have taken him if there had been a perfect calm? Of course, the hydroplane's engine worked uniformly throughout. (3/3)
Tommy replied, "it is true that in Ireland there are men of Cork and in Scotland men of Ayr, which is better still, but in England there are lightermen." Unfortunately it had to be explained to Mrs. Dobson, and this took the edge off the thing. The hydroplane flight was from Slocomb to the neighbouring watering-place Poodleville--five miles distant. But there was a strong wind, which so helped the airman that he made the (2/3)
72. THE HYDROPLANE QUESTION. The inhabitants of Slocomb-on-Sea were greatly excited over the visit of a certain flying man. All the town turned out to see the flight of the wonderful hydroplane, and, of course, Dobson and his family were there. Master Tommy was in good form, and informed his father that Englishmen made better airmen than Scotsmen and Irishmen because they are not so heavy. "How do you make that out?" asked Mr. Dobson. "Well, you see," (1/3)
SOLUTION TO 396. A MATCH MYSTERY. (4/4)
And so on to the end. This solution is quite general, and applies to any number of matches and any number of heaps. A correspondent informs me that this puzzle game was first propounded by Mr. W.M.F. Mellor, but when or where it was published I have not been able to ascertain.
SOLUTION TO 396. A MATCH MYSTERY. (3/4)
As there are thus two of every power, you must win. Say your opponent takes 7 from the 12 heap. He then leaves--
5 = - 4 - 1
11 = 8 - 2 1
7 = - 4 2 1
1 2 2 3
Here the powers are not all even in number, but by taking 9 from the 11 heap you immediately restore your winning position, thus--
5 = - 4 - 1
2 = - - 2 -
7 = - 4 2 1
- 2 2 2
SOLUTION TO 396. A MATCH MYSTERY. (2/4)
remembering that 2^0 = 1. Then if you so leave the matches to your opponent that there is an even number of every power, you can win. And if at the start you leave the powers even, you can always continue to do so throughout the game. Take, as example, the last grouping given above--12, 11, 7. Expressed in powers of 2 we have--
12 = 8 4 - -
11 = 8 - 2 1
7 = - 4 2 1
2 2 2 2
SOLUTION TO 396. A MATCH MYSTERY. (1/4)
If you form the three heaps (and are therefore the second to draw), any one of the following thirteen groupings will give you a win if you play correctly: 15, 14, 1; 15, 13, 2; 15, 12, 3; 15, 11, 4; 15, 10, 5; 15, 9,
6; 15, 8, 7; 14, 13, 3; 14, 11, 5; 14, 9, 7; 13, 11, 6; 13, 10, 7; 12,
The beautiful general solution of this problem is as follows. Express the number in every heap in powers of 2, avoiding repetitions and
matches may be grouped at the start for a certain win. In fact, the groups selected, 14, 11, 5, are a certain win, because for whatever your opponent may play there is another winning group you can secure, and so on and on down to the last match." (5/5)
Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3;
Mr. Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr. Stubbs, 2, 3, 1; Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1,
"It is now quite clear that I must win," said Mr. Stubbs, because you must take 1, and then I take 1, leaving you the last match. You never had a chance. There are just thirteen different ways in which the (4/5)
in one heap I can repeat in the other. If you leave 4 in one heap, I leave 4 in the other. If you then leave 2 in one heap, I leave 2 in the other. If you leave only 1 in one heap, then I take all the other heap. If you take all one heap, I take all but one in the other. No, you must never leave two heaps, unless they are equal heaps and more than 1, 1. Let's begin again."
"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and leave you 8, 11, 5."
the last match loses the game. That's all! I will play with you, Wilson. I have formed the heaps, so you have the first draw."
"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usual moderation and take all the 14 heap."
"That is the worst you could do, for it loses right away. I take 6 from the 11, leaving two equal heaps of 5, and to leave two equal heaps is a certain win (with the single exception of 1, 1), because whatever you do (2/5)
396. A MATCH MYSTERY. Here is a little game that is childishly simple in its conditions. But it is well worth investigation.
Mr. Stubbs pulled a small table between himself and his friend, Mr. Wilson, and took a box of matches, from which he counted out thirty.
"Here are thirty matches," he said. "I divide them into three unequal heaps. Let me see. We have 14, 11, and 5, as it happens. Now, the two players draw alternately any number from any one heap, and he who draws (1/5)
SOLUTION TO 394. PUSS IN THE CORNER. (2/2)
33, if he has first move, he prevents B getting diagonally in line with himself. Here are two good games. The number in front of the hyphen is always A's move; that after the hyphen is B's:--
33-8, 32-15, 31-22, 30-21, 29-14, 22-7, 15-6, 14-2, 7-3, 6-4, 11-, and A must capture on his next (12th) move, -13, 54-20, 53-27, 52-34, 51-41,
50-34, 42-27, 35-20, 28-13, 21-6, 14-2, 7-3, 6-4, 11-, and A must capture on his next (14th) move.
SOLUTION TO 394. PUSS IN THE CORNER. (1/2)
No matter whether he plays first or second, the player A, who starts the game at 55, must win. Assuming that B adopts the very best lines of play in order to prolong as much as possible his existence, A, if he has first move, can always on his 12th move capture B; and if he has the second move, A can always on his 14th move make the capture. His point is always to get diagonally in line with his opponent, and by going to
15; A retreats to 26; B retreats to 13; A advances to 21; B retreats to 2; A advances to 7; B goes to 3; A moves to 6; B must now go to 4; A establishes himself at 11, and B must be captured next move because he is compelled to cross a line on which A stands. Play this over and you will understand the game directly. Now, the puzzle part of the game is this: Which player should win, and how many moves are necessary? (2/2)
394. PUSS IN THE CORNER. This variation of the last puzzle is also played by two persons. One puts a counter on No. 6, and the other puts one on No. 55, and they play alternately by removing the counter to any other number in a line. If your opponent moves at any time on to one of the lines you occupy, or even crosses one of your lines, you immediately capture him and win. We will take an illustrative game.
A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to (1/2)
SOLUTION TO 107. THE SUFFRAGISTS' MEETING.
Eighteen were present at the meeting and eleven left. If twelve had gone, two-thirds would have retired. If only nine had gone, the meeting would have lost half its members.
Can you tell how many were present at the meeting at the start? (2/2)
107. THE SUFFRAGISTS' MEETING. At a recent secret meeting of Suffragists a serious difference of opinion arose. This led to a split, and a certain number left the meeting. "I had half a mind to go myself," said the chair-woman, "and if I had done so, two-thirds of us would have retired." "True," said another member; "but if I had persuaded my friends Mrs. Wild and Christine Armstrong to remain we should only have lost half our number." (1/2)
SOLUTION TO 357. ANCIENT CHINESE PUZZLE.
Play as follows:--
1. R--Q 6
2. K--R 7
3. R (R 6)--B 6 (mate).
Black's moves are forced, so need not be given.