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.
SOLUTION TO 394. PUSS IN THE CORNER. (2/2) Show more
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) Show more
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 95. THE FOUR SEVENS. Show more
The way to write four sevens with simple arithmetical signs so that they represent 100 is as follows:--
-- × -- = 100.
Of course the fraction, 7 over decimal 7, equals 7 divided by 7/10,
which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10 is 100, and there you are! It will be seen that this solution applies equally to any number whatever that you may substitute for 7.
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
or 48 yards 3 inches, is the required length of the tether "to the nearest inch."
SOLUTION TO 196. THE TETHERED GOAT. (2/3) Show more
require is the radius (CD) of a circle containing six quarter-acres or 1½ acres, which is equal to 9,408,960 square inches. As we only want our answer "to the nearest inch," it is sufficiently exact for our purpose if we assume that as 1 is to 3.1416, so is the diameter of a circle to its circumference. If, therefore, we divide the last number I gave by 3.1416, and extract the square root, we find that 1,731 inches,
SOLUTION TO 196. THE TETHERED GOAT. (1/3) Show more
This problem is quite simple if properly attacked. Let us suppose the triangle ABC to represent our half-acre field, and the shaded portion to be the quarter-acre over which the goat will graze when tethered to the corner C. Now, as six equal equilateral triangles placed together will form a regular hexagon, as shown, it is evident that the shaded pasture is just one-sixth of the complete area of a circle. Therefore all we
SOLUTION TO 64. THE THREE CLOCKS. (4/4) Show more
that 720 days from noon of April 1, 1898, brings us to noon of March 22,
SOLUTION TO 64. THE THREE CLOCKS. (3/4) Show more
four without a remainder is bissextile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year,
nor was 1900. On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered bissextile. Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. We therefore find
SOLUTION TO 64. THE THREE CLOCKS. (2/4) Show more
other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?
I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by
SOLUTION TO 64. THE THREE CLOCKS. (1/4) Show more
As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time,
it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours,
and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the
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
18-9). When you have made the interchanges within any pair of brackets,
all numbers within those brackets are in their places. There are five pairs of brackets, and 5 from 22 gives the number of changes required--17.
SOLUTION TO 238. ARRANGING THE JAM POTS. (1/2) Show more
Two of the pots, 13 and 19, were in their proper places. As every interchange may result in a pot being put in its place, it is clear that twenty-two interchanges will get them all in order. But this number of moves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17),
(24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14,
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
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