143. TWO CROSSES FROM ONE. Cut a Greek cross into five pieces that will form two such crosses, both of the same size. The solution of this puzzle is very beautiful.
SOLUTION TO 296. THE FOUR LIONS. Show more
There are only seven different ways under the conditions. They are as follows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1 3. Taking the last example, this notation means that we place a lion in the second square of first row, fourth square of second row, first square of third row, and third square of fourth row. The first example is, of course, the one we gave when setting the puzzle.
you merely get the first arrangement. It is a simple little puzzle, but requires a certain amount of careful consideration. (2/2)
296. THE FOUR LIONS. The puzzle is to find in how many different ways the four lions may be placed so that there shall never be more than one lion in any row or column. Mere reversals and reflections will not count as different. Thus, regarding the example given, if we place the lions in the other diagonal, it will be considered the same arrangement. For if you hold the second arrangement in front of a mirror or give it a quarter turn, (1/2)
inconveniently, you cannot obtain a solution by breaking a Queen's Tour,
or in any other way by queen moves alone. But you are allowed to use oblique straight lines--such as from the upper white star direct to a corner star. (2/2)
329. THE STAR PUZZLE. Put the point of your pencil on one of the white stars and (without ever lifting your pencil from the paper) strike out all the stars in fourteen continuous straight strokes, ending at the second white star. Your straight strokes may be in any direction you like, only every turning must be made on a star. There is no objection to striking out any star more than once.
In this case, where both your starting and ending squares are fixed (1/2)
SOLUTION TO 38. THE BICYCLE THIEF. (2/2) Show more
and the other operations of changing the cheque and borrowing from a friend do not affect the question in the slightest. The loss of prospective profit on the sale of the bicycle is, of course, not direct loss of money out of pocket.
SOLUTION TO 38. THE BICYCLE THIEF. (1/2) Show more
People give all sorts of absurd answers to this question, and yet it is perfectly simple if one just considers that the salesman cannot possibly have lost more than the cyclist actually stole. The latter rode away with a bicycle which cost the salesman eleven pounds, and the ten pounds
"change;" he thus made off with twenty-one pounds, in exchange for a worthless bit of paper. This is the exact amount of the salesman's loss,
amount he had received. To do this, he was compelled to borrow the £25 from a friend, as the cyclist forgot to leave his address, and could not be found. Now, as the bicycle cost the salesman £11, how much money did he lose altogether? (2/2)
38. THE BICYCLE THIEF. Here is a little tangle that is perpetually cropping up in various guises. A cyclist bought a bicycle for £15 and gave in payment a cheque for £25. The seller went to a neighbouring shopkeeper and got him to change the cheque for him, and the cyclist, having received his £10 change, mounted the machine and disappeared. The cheque proved to be valueless, and the salesman was requested by his neighbour to refund the (1/2)
SOLUTION TO 258. THE VOTERS' PUZZLE. Show more
THE number of readings here is 63,504, as in the case of "WAS IT A RAT I SAW" (No. 30, _Canterbury Puzzles_). The general formula is that for palindromic sentences containing 2n + 1 letters there are (4(2^n -1))² readings.
SOLUTION TO 60. THE WAPSHAW'S WHARF MYSTERY. (3/3) Show more
the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. 49+1/11 sec. out in his reckoning.
SOLUTION TO 60. THE WAPSHAW'S WHARF MYSTERY. (2/3) Show more
(twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4 hr. 21 min. 49+1/11 sec. This last is the only occasion on which the two hands are together with the second hand "just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his _Across the World for a Wife_, says, "It was a cold, dreary winter's afternoon, and by the time
SOLUTION TO 60. THE WAPSHAW'S WHARF MYSTERY. (1/3) Show more
There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide 12 hours by 11 we get 1 hr. 5 min. 27+3/11 sec., and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54+6/11 sec.
recollected that the hour hand and minute hand were exactly together,
one above the other, and the second hand had just passed the forty-ninth second. More than this he could not remember.
What was the exact time at which the watchman's watch stopped? The watch is, of course, assumed to have been an accurate one. (4/4)
But a very stupid officer (and we invariably find one or two stupid individuals in the most intelligent bodies of men) had actually amused himself by turning the hands round and round, trying to set the watch going again. After he had been severely reprimanded for this serious indiscretion, he was asked whether he could remember the time that was indicated by the watch when found. He replied that he could not, but he (3/4)
the robbery. In this belief the proprietors were confirmed when, later in the day, they were informed that the poor fellow's body had been picked up by the River Police. Certain marks of violence pointed to the fact that he had been brutally attacked and thrown into the river. A watch found in his pocket had stopped, as is invariably the case in such circumstances, and this was a valuable clue to the time of the outrage. (2/4)
60. THE WAPSHAW'S WHARF MYSTERY. There was a great commotion in Lower Thames Street on the morning of January 12, 1887. When the early members of the staff arrived at Wapshaw's Wharf they found that the safe had been broken open, a considerable sum of money removed, and the offices left in great disorder. The night watchman was nowhere to be found, but nobody who had been acquainted with him for one moment suspected him to be guilty of (1/4)
SOLUTION TO 25. CHINESE MONEY. (2/2) Show more
worth one-eleventh of 16 ching-changs--that is, 77 rounds equal 105 ching-changs and 11 squares equal 16 ching-changs. Therefore 77 rounds added to 11 squares equal 121 ching-changs; or 7 rounds and 1 square equal 11 ching-changs, or its equivalent, half a crown. This is more simple in practice than it looks here.