...
If you simply turn the cards round so that one of the other two sides is nearest to you this will not count as different, for the order will be the same. Also, if you make the 4, 9, 5 change places with the 7, 3, 8,
and at the same time exchange the 1 and the 6, it will not be different. But if you only change the 1 and the 6 it will be different, because the order round the triangle is not the same. This explanation will prevent any doubt arising as to the conditions. (2/2)

384. CARD TRIANGLES. Here you pick out the nine cards, ace to nine of diamonds, and arrange them in the form of a triangle, exactly as shown in the illustration, so that the pips add up the same on the three sides. In the example given it will be seen that they sum to 20 on each side, but the particular number is of no importance so long as it is the same on all three sides. The puzzle is to find out in just how many different ways this can be done.
(1/2)

SOLUTION TO 394. PUSS IN THE CORNER. (2/2) Show more

SOLUTION TO 394. PUSS IN THE CORNER. (1/2) Show more

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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 95. THE FOUR SEVENS. Show more

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represent 100. Every juvenile reader will see at a glance that his example is quite correct. Now, what he wants you to do is this: Arrange four 7's (neither more nor less) with arithmetical signs so that they shall represent 100. If he had said we were to use four 9's we might at once have written 99+9/9, but the four 7's call for rather more ingenuity. Can you discover the little trick? (2/2)

95. THE FOUR SEVENS. In the illustration Professor Rackbrane is seen demonstrating one of the little posers with which he is accustomed to entertain his class. He believes that by taking his pupils off the beaten tracks he is the better able to secure their attention, and to induce original and ingenious methods of thought. He has, it will be seen, just shown how four 5's may be written with simple arithmetical signs so as to (1/2)

SOLUTION TO 196. THE TETHERED GOAT. (3/3) Show more

SOLUTION TO 196. THE TETHERED GOAT. (2/3) Show more

SOLUTION TO 196. THE TETHERED GOAT. (1/3) Show more

SOLUTION TO 64. THE THREE CLOCKS. (4/4) Show more

SOLUTION TO 64. THE THREE CLOCKS. (3/4) Show more

SOLUTION TO 64. THE THREE CLOCKS. (2/4) Show more

SOLUTION TO 64. THE THREE CLOCKS. (1/4) Show more

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progress without stopping, on what date and at what time of day would all three pairs of hands again point at the same moment at twelve o'clock? (2/2)

64. THE THREE CLOCKS. On Friday, April 1, 1898, three new clocks were all set going precisely at the same time--twelve noon. At noon on the following day it was found that clock A had kept perfect time, that clock B had gained exactly one minute, and that clock C had lost exactly one minute. Now, supposing that the clocks B and C had not been regulated, but all three allowed to go on as they had begun, and that they maintained the same rates of (1/2)

SOLUTION TO 238. ARRANGING THE JAM POTS. (2/2) Show more

SOLUTION TO 238. ARRANGING THE JAM POTS. (1/2) Show more

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