contain the same quantity of fluid. What proportion of spirits to water did the spirits of wine bottle then contain?" (2/2)
363. THE DOCTOR'S QUERY.
"A curious little point occurred to me in my dispensary this morning,"
said a doctor. "I had a bottle containing ten ounces of spirits of wine,
and another bottle containing ten ounces of water. I poured a quarter of an ounce of spirits into the water and shook them up together. The mixture was then clearly forty to one. Then I poured back a quarter-ounce of the mixture, so that the two bottles should again each (1/2)
SOLUTION TO 69. THE THREE VILLAGES. (2/2) Show more
figures are easily proved, for the square of 12 added to the square of 9 equals the square of 15, and the square of 12 added to the square of 16 equals the square of 20.
SOLUTION TO 69. THE THREE VILLAGES. (1/2) Show more
Calling the three villages by their initial letters, it is clear that the three roads form a triangle, A, B, C, with a perpendicular,
measuring twelve miles, dropped from C to the base A, B. This divides our triangle into two right-angled triangles with a twelve-mile side in common. It is then found that the distance from A to C is 15 miles, from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These
69. THE THREE VILLAGES. I set out the other day to ride in a motor-car from Acrefield to Butterford, but by mistake I took the road going _via_ Cheesebury, which is nearer Acrefield than Butterford, and is twelve miles to the left of the direct road I should have travelled. After arriving at Butterford I found that I had gone thirty-five miles. What are the three distances between these villages, each being a whole number of miles? I may mention that the three roads are quite straight.
SOLUTION TO 236. THE HAT PUZZLE. (2/2) Show more
11, and, finally, 1 and 2, leaving the four silk hats together, the four felt hats together, and the two vacant pegs at one end of the row. The first three pairs moved are dissimilar hats, the last two pairs being similar. There are other ways of solving the puzzle.
SOLUTION TO 236. THE HAT PUZZLE. (1/2) Show more
I suggested that the reader should try this puzzle with counters, so I give my solution in that form. The silk hats are represented by black counters and the felt hats by white counters. The first row shows the hats in their original positions, and then each successive row shows how they appear after one of the five manipulations. It will thus be seen that we first move hats 2 and 3, then 7 and 8, then 4 and 5, then 10 and
SOLUTION TO 400. THE MAGIC STRIPS. (3/3) Show more
be moved in succession from left to right or from right to left, each time producing a magic square.
SOLUTION TO 400. THE MAGIC STRIPS. (2/3) Show more
that the uncut strip is at the top, but it will be found that if the bottom row of figures be placed at the top the numbers will still form a magic square, and that every successive removal from the bottom to the top (carrying the uncut strip stage by stage to the bottom) will produce the same result. If we imagine the numbers to be on seven complete _perpendicular_ strips, it will be found that these columns could also
SOLUTION TO 400. THE MAGIC STRIPS. (1/3) Show more
There are of course six different places between the seven figures in which a cut may be made, and the secret lies in keeping one strip intact and cutting each of the other six in a different place. After the cuts have been made there are a large number of ways in which the thirteen pieces may be placed together so as to form a magic square. Here is one of them:--
The arrangement has some rather interesting features. It will be seen
piece containing a number, and the puzzle would then be very easy, but I need hardly say that forty-nine pieces is a long way from being the fewest possible. (3/3)
the numbers shall form seven rows and seven columns.
Now, the puzzle is to cut these strips into the fewest possible pieces so that they may be placed together and form a magic square, the seven rows, seven columns, and two diagonals adding up the same number. No figures may be turned upside down or placed on their sides--that is, all the strips must lie in their original direction.
Of course you could cut each strip into seven separate pieces, each (2/3)
400. THE MAGIC STRIPS. I happened to have lying on my table a number of strips of cardboard,
with numbers printed on them from 1 upwards in numerical order. The idea suddenly came to me, as ideas have a way of unexpectedly coming, to make a little puzzle of this. I wonder whether many readers will arrive at the same solution that I did.
Take seven strips of cardboard and lay them together as above. Then write on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so that (1/3)
SOLUTION TO 172. MRS. SMILEY'S CHRISTMAS PRESENT. (2/2) Show more
will fit together and form a square in each case. The assembling of the parts may be slightly varied, and the reader may be interested in finding a solution for the third set of squares I have given.
SOLUTION TO 172. MRS. SMILEY'S CHRISTMAS PRESENT. (1/2) Show more
The first step is to find six different square numbers that sum to 196. For example, 1 + 4 + 25 + 36 + 49 + 81 = 196; 1 + 4 + 9 + 25 + 36 + 121
= 196; 1 + 9 + 16 + 25 + 64 + 81 = 196. The rest calls for individual judgment and ingenuity, and no definite rules can be given for procedure. The annexed diagrams will show solutions for the first two cases stated. Of course the three pieces marked A and those marked B
SOLUTION TO 427. PHEASANT-SHOOTING. (2/2) Show more
shot remained, because the others flew away? No; because the question was not "how many remained?" but "how many still remained?" Now the poor bird that was wounded in the wing, though unable to fly, was very active in its painful struggles to run away. The answer is, therefore, that the 16 birds that were shot dead "still remained," or "remained still."
SOLUTION TO 427. PHEASANT-SHOOTING. (1/2) Show more
The arithmetic of this puzzle is very easy indeed. There were clearly 24 pheasants at the start. Of these 16 were shot dead, 1 was wounded in the wing, and 7 got away. The reader may have concluded that the answer is,
therefore, that "seven remained." But as they flew away it is clearly absurd to say that they "remained." Had they done so they would certainly have been killed. Must we then conclude that the 17 that were
wing right in front of us. I fired, and two-thirds of them dropped dead at my feet. Then the duke had a shot at what were left, and brought down three-twenty-fourths of them, wounded in the wing. Now, out of those twenty-four birds, how many still remained?"
It seems a simple enough question, but can the reader give a correct answer? (2/2)
427. PHEASANT-SHOOTING. A Cockney friend, who is very apt to draw the long bow, and is evidently less of a sportsman than he pretends to be, relates to me the following not very credible yarn:--
"I've just been pheasant-shooting with my friend the duke. We had splendid sport, and I made some wonderful shots. What do you think of this, for instance? Perhaps you can twist it into a puzzle. The duke and I were crossing a field when suddenly twenty-four pheasants rose on the (1/2)
SOLUTION TO 139. THE DUTCHMEN'S WIVES. (8/8) Show more
may be expressed as the difference of two squares in more than one way,
or proved to be a prime. But the method, when dealing with large numbers, is necessarily tedious, though in practice it may be considerably shortened. In many cases it is the shortest method known for factorizing large numbers, and I have always held the opinion that Fermat used it in performing a certain feat in factorizing that is historical and wrapped in mystery.
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
A Mastodon instance for maths people. The kind of people who make \(\pi z^2 \times a\) jokes.
\) for inline LaTeX, and
\] for display mode.