Tag Archives: puzzle

Barrels, Cans, Flasks, Glasses, Tanks and Vessels pouring puzzles


1) 3, 4 and 5*
My grandfather was a milkman in the time when milkmen went from door to door, and the women came out with a can to buy some milk. His car held a large tank with tens of liters of milk.
Normally my granddad had a number of measures with him, but one day he had just a 3 quarter can with him. One woman came out with her 5 quarter can and wanted to purchase exactly 4 quarter.
How did my grandfather arrange this? Of course no milk should be spilled.

If you solved it, you can check your solution.

2) 3, 5 and 8 – divide equally*
Sam Loyd mentions the puzzle above as a very old problem. Actually, this is a simplified form of an old puzzle, popularized by Niccolo Fontana (1500-1559), nicknamed the Stammerer (Tartaglia), which ran:

There are three jugs A, B, C, with capacities 3, 5, and 8 quarts, respectively. Jug C is filled with wine, and we wish to divide the wine into two equal parts by pouring it from one container to another – that is, without using any measuring devices other than these jugs.

Fontana probably got it from an older manuscript dated around 1200.

This puzzle was discussed mathematically by Alexander Bogomolny

If you solved it, you can check your solution

.

3) Three glasses puzzle*
There are three glasses on the table – 3, 5, and 8 oz. The first two are empty, the last contains 8 oz of water. By pouring water from one glass to another make at least one of them contain exactly 4 oz of water.
This is a waterdown version of the previous puzzle, so we’ll skip this and instead propose a version described by puzzle blogger Professor Richard Wiseman:

There are 3 glasses of 3, 5 and 8 oz. The 3 oz glass holds 2 oz, the 5 oz glass holds 3 oz and the 8 oz glass holds 5 oz.
Can you measure 1 oz in only 2 pourings?
solution

4) Divide 24 liters in three equal portions*
Divide 24 fluid ounces of a liquid into three equal parts, given that there are three containers available holding 5, 11 and 13 fluid ounces.
This problem was posed by Claude Gaspard Bachet de Méziriac in his Problèmes plaisans et délectables (1612). He may have copied this problem from an older resource such as Tartaglia.
Henry Dudeney remarked about this puzzle: There are many different solutions to this puzzle in six manipulations, or pourings from one vessel to another.
solution

In the second half of the 19th century, both Sam Loyd and Henry Dudeney produced puzzles of this type:
5) The Wassail bowl*
One Christmas Eve three Weary Willies came into possession of what was to them a veritable wassail bowl, in the form of a small barrel, containing exactly six quarts of fine ale. One of the men possessed a five-pint jug and another a three-pint jug, and the problem for them was to divide the liquor equally amongst them without waste. Of course, they are not to use any other vessels or measures. If you can show how it was to be done at all, then try to find the way that requires the fewest possible manipulations, every separate pouring from one vessel to another, or down a man’s throat, counting as a manipulation.
solution

6) The barrel puzzle*

The men in the illustration are disputing over the liquid contents of a barrel. What the particular liquid is it is impossible to say, for we are unable to look into the barrel; so we will call it water. One man says that the barrel is more than half full, while the other insists that it is not half full. What is their easiest way of settling the point? It is not necessary to use stick, string, or implement of any kind for measuring. I give this merely as one of the simplest possible examples of the value of ordinary sagacity in the solving of puzzles. What are apparently very difficult problems may frequently be solved in a similarly easy manner if we only use a little common sense.
This puzzle stems from Dudeneys “Amusement in mathematics”
solution

7) 4, 5, 10 and 10**
Here is a new poser (wrote Henry Dudeney at the end of the 19th century) in measuring liquids that will be found interesting. A man has two ten-quart vessels full of wine, and a five-quart and a four-quart measure. He wants to put exactly three quarts into each of the two measures. How is he to do it? And how many manipulations (pourings from one vessel to another) do you require? Of course, waste of wine, tilting, and other tricks are not allowed.
This puzzle stems from Dudeneys “Amusement in mathematics”
solution

8) The brook*
Another Dudeney: A man goes to a brook with two measures of 15 and 16 pints. How is he to measure exactly 8 pints in the fewest possible moves?
I need hardly add that no tricks such as tilting or marking are allowed.
solution

9) The barrels of beer*
The america prohibition authority discovered a full barrel of beer, and were about to destroy the liquid by letting it run down a drain when the owner the owner pointed to 2 vessels standing by and begged to be allowed to retain in them a small quantity for the immedeat consumption of his household. One vessel was a 7 quart and the other a 5 quart measure. The officer was a wag, and believing it to be impossible, said that if the man could measure an exact quart into each vessel, without pouring anything back into the barrel, he might do so.
How was it to be done in the fewest possible transactions without any tricks such as marking?
Perhaps I should add that an american barrel of beer contains exactly 120 quarts. Wasting beer is in this instance of course allowed.
solution

10) The merchant of Bagdad*

Sam Loyd presented the following problem on page 188 of his “Cyclopedia of Puzzles”.
Ae merchant of Baghdad who catered to the wants of pilgrims who ncrossed the desert, was once confronted by the following perplexing problem:

He was visited by the leader of of a caravan, who desired to purchase a store of wine and water. Presenting three 10 gallon vessels, he asked that three gallons of wine be put in the first, three gallons of water in the second, and three of wine and three of water mixed in the third, and that three gallons of water be given to each of the thirteen camels.

As both water and wine, according to Oriental usage, are only sold in quantities of an even number of gallons, the merchant had only a two and a four gallen measure wherewith to perform a feat which presents some unexpected difficulties; nevertheless, without resorting to any trick or device, or expedient not pertaining to the ordinary measuring problem, as already referred to, he dispensed the water from a full hoggshead, and the wine from a barrel, in the required proportions, without any waste whatever.
In how few manipulations can the feat be performed, counting every time that liquid is drawn from one receptacle to another as a manipulation?

Sam Loys son, Sam Loyd jr, added: This puzzle is undoubtedly the most remarkable problem of its extant, anf for many years baffled the puzzlists of the world to reduce to the least possuiible number of “moves”, as the manipulations were then termed. By many it has been referred to as Sam Loyds greatest puzzle.
From a contemporary view, it may be useful to note that a hoggshead contains 63 gallons, while a barrel contains 31.5 gallons.
solution

This one too comes from Sam Loyd:
11) Milkmans puzzle**

There are practical problems in all trades, so it is safe to say that no one is an adept at his business unless he had pickesd up a few wrinkles which pertain to his calling.

Honest John says that what he “dont know about milk is scarcely worth mentioning”, but he was nearly flabbergasted when he had two 10 gallon cans full of milk and two customers with a four and a five quarter measure wanted two quarts in each measure.

It is a juggling trick pure and simple, devoid of trick or device, but it calls for much cleverness to get two exact quarts of milk in those two measures employing no receptacles of any kind except the two measures and the two full cans. You cab try the problem with the fullest assurance that it is a legitimate problem and not a silly catch.

According to professor Knuths index, this puzzle was first published in Women’s Home Companion 34,3 (March 1907).
solution

12) Three glasses puzzle – again**

A variant of my own:
You have three glasses with a capacity of 3, 5 and 8 oz, partly filled with water.
The 3oz glass holds 2oz.
The 5oz glass holds 3oz.
The 8oz glass holds 5oz.
If spilling water is not considered a move, how can you measure 1 oz in 1 move?
solution

Soma cube – part 2


Some time ago I presented some new soma cube puzzles. If you dont know what a soma cube is, have a look here.

Here are 8 new figures:
1) wall
soma cube exercise wall 09


2) stairs 1
soma cube exercise stairs 10


3) airplane
soma cube exercise plane 11


4) stairs 2
soma cube exercise stairs 12


5) castle wall 1
soma cube exercise castle wall 13


6) gate
soma cube exercise gate 14


7) castle tower 2
soma cube exercise castle tower 15


8) castle tower 3
soma cube exercise castle tower 16


If you solved it, we have the solution to
wall,
stairs 1,
airplane,
stairs 2,
castle wall 1,
castle gate,
castle tower 2,
castle tower 3,

Camel inheritance



1) How many camels?*
The sheik has died. When the Mullah read the will, he found that the sheik had left each of his five sons 1/6 of his camel herd, while his only daughter in an act of sheer discrimination inherited only 1/8th of the herd.
The mullah solved it for the kids without butchering a camel.
How many camels did the sheik have?

You can check your solution.

The puzzle above is a new one, and of course derived from the following classic:
2) 17 camels and three sons**
The sheik has passed away. When the mullah opens his will, he finds the sheik has left 1/2 of his camels to his oldest son, Achmed, 1/3 to the second son, Harim, and 1/9th to poor Bahari, the youngest. Now one of the sheiks camels had died in an accident a month ago, leaving only 17 camels to be divided.
How did the mullah divide the camels without butchering one?

This puzzle is based on a problem which according to some was first posed by Gaston Boucheny, “Curiosités et Récréations Mathématiques”. Paris, 1939. The French ed. of MRE says it is a problem of Arabic origin, while Kraitchik, Math. des Jeux, says it is a Hindu problem. The claim attributing the puzzle to 1939 seems wrong to me, as Sam Loyd and Henry Dudeney posed the problem in the Strand magazine years earlier. In fact, this puzzle was included in Henry’s puzzle book: “536 Puzzles and Curious problems” as number 172.

There is a hint.

3) The seventeen horses**
“I suppose you all know this old puzzle” said Jeffries. “A farmer left his seventeen horses to be divided among his three sons in the following proportions: one half to the eldest, one third to the second, and one ninth to the youngest. How should they be divided?”
“Yes, we all know that”, said Robinson. “But it’s impossible. The answer given is always a fallacy.”
“I suppose you mean,” Progers suggested, “the answer where one horse is borrowed, so that the division can be done without butchering a horse, the sons receive9, 6 and 2 and the extra horse is returned”.
“Exactly!” Robinson replied “And each son receives more than his share.”
“Stop!” cried Benson. “If each man receives more than his share, the total must exceed 17 horses, but 9, 6 and 2 neatly sum up to 17.”
“That indeed looks queer”, Robinson admitted, “but 17/2 is 8,5, not 9. so the oldest son receives more than his share. And it’s similar for the other sons. The thing can’t really been done”
“And that’s where you all are wrong”. Jeffries stated. “The terms of the will can exactly be carried out, without any mutilation of a horse.”
To their astonishment, he showed them how it was possible.

There is a hint.

Oh, the image at the top of this page is the coat of arms of Zurich, available under GFDL license and created by Ronald zh.