Category Archives: Mathematics

Review crystal jigsaw puzzles


This is one of three reviews to be published this summer while I’m away on holiday,

This post is about plastic 3D jigsaw puzzles that I received for my birthday and fathersday. My thanks of course go to the family members who donated them.

Now I must confess that I dont like jigsaws puzzles. One reasons is they are way too common for my taste. Another reason is that they do not tax my brain enough: it needs diligent work, but not hard work.
But when it comes to 3D puzzles, the borderline becomes fuzzy. Ravensburg has published 3D jigsaw puzzles for I estimate over a decade. All pieces look like ordinary pieces, except they are thicker, so you can build walls and roofs with them.

On the other end of the 3d spectrum are the Japanese kumiki puzles, often beautifully made of wood, and representing several types of buildings, animals, fruit and other objects.

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Somewhere in between are the plastic 3D jigsaw puzzles. I discovered them on the web, as produced by Kimzel Gmbh in Germany. They have a series of about 40 models. I ordered mine through Moenen and Mariken. This shop is in Dutch, and I can recommend their service, which was excellent.
If you live in the USA, you can order them from Amazon, where they are marketed by BePuzzled.

The number of pieces varies greatly. I have an apple of 13 pices, several of 46 pieces, and I have seen some of 90-100 pieces. Despite their name, they are not really crystal, but a sturdy plastic that looks like it. The copyright belongs to Beverley Enterprises Inc. & Jeruel Ind Co Ltd. The latter is a China based Toys and Puzzle producer, and the puzzles are produced in China. The pieces fit together really well. I guess the plastic does make them affordable, prices range from 3-20 euros.

Review Hanayama puzzles


This is one of three reviews to be published this summer while i’m away on holiday,

This post is about metal puzzles that I received for my birthday and fathersday. My thanks of course go to the family members who donated them.

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The Hanayama series is a series of beautiful iron cast puzzles published by hanayame toys in Japan.

There are 2 kinds of packaging around, one is a beautiful shiny red/black cardboard box, the other a dull grey-green? cardboard box, which I dont have myself but did see on ads in the USA.

The puzzles themselves are very solid, and have varying degrees of difficulty. The manufacturer has them labeled with a number of stars, more stars meaning more difficult. Many of the puzzles have been designed by well established names in the puzzle world such as Oskar van Deventer and Nob Yoshigahara.

New puzzles seem to be published regularly, and can be mail ordered at several places:

They have varying degrees of difficulty, 1-5, and come without a solution, though you may try the generally know video sites if you are completely stuck. I now have one of the rings, the cast duet and the cast quartet, and probably one or more others, and I’m very satisfied with them. There must be somewhere between 50 and a 100 of them by now,

Strange calculation?


Intro whats next numbers

 

 
IF

8 x 2 = 46416
6 + 3 = 23618
15 : 3 = 513545
9 – 3 = 38127

Then what is :
12 x 4 =?

I am very much interested in your solution times, and you are welcome to make remarks, and discuss alternatives. Pointing out alternative solutions is also welcome, as they point out possible problems in the brain teasers, but please dont mention solutions and leave others the fun to solve them too.

If you are puzzled, we have a solution for you.

81


Carte_française_trèfle_04

Carte_française_pique_04

Carte_française_cœur_04

Carte_française_carreau_04


Using four times the number four, create the number 81. You may use the usual mathematical operands.

If you are puzzled, we have a solution for you.

I would like to thank Kees Krol again for coming up with this one.

As always, please don’t publish your solutions. Solutions can be found after 1-2 weeks on the solution page for those who want to check their solutions, or for those who are really stuck.

But scrolling is much easier, and really spoils the fun for others.

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

Socks – again


Socks
In one of the earliest posts I listed a number of puzzles on socks. Here is another one:

1) Socks in a bag
You put two socks in a bag. For every sock you put in, there is a 50% chance that it is blue and a 50% chance that it is red. I don’t see which colours you put into the bag.
Without looking, I take out one sock and it turns out to be blue. What is the chance that the other is also blue?

If you are puzzled, we have a solution for you.

Make 21


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82px-05_of_diamonds

82px-06_of_hearts

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1)Make 21*
No, it’s not the card game.

The puzzle is can simply be formulated: make 21 using the numbers 1, 5, 6 and 7 once each, by combining them with the operators + – * / and the ().
The word ‘numbers’ is on purpose, i do mean numbers and not digits. And there are no tricks involved such as turning digits around, making a 9 from the 6.

Kees Krol also brought this puzzle to my attention, again: Thank you, Kees!

You can check your solution

Rowmix


One puzzle type occasionally appearing in Visie, a magazine of the Evangelical Broadcast in the Netherlands, is something the author calls “All mixed up”. I prefer “Rowmix” for reasons which will be clear once you have take a look at the puzzle rules.

Rowmix nr 1*

5 6 7 8
1 /2 +7 -5 x5 15
2 x4 -1 +3 /2 12
3 +8 +3 /2 -1 8
4 +7 +7 -6 -2 10
12 8 15 22

Shuffle the numbers in every row so that a correct calculation is formed in every row and column. In doing so, all calculations are performed strictly from left to right or from top to bottom. There is no precedence of multiplication and division over addition and subtraction, and neither are there any brackets to be inserted.

For example: In the first row above you should, starting with 1, using /2, +7, -5, and x5, make 15.

Rowmix nr 2*

12 12 12 12
10 /2 -3 +4 +1 5
10 x2 /2 +5 -10 10
10 +10 -9 -5 -1 5
10 +14 +6 +3 +2 35
24 12 12 12

Rowmix nr 3*

1 2 3 4
1 x3 /2 +8 +7 20
2 /4 +6 +2 +2 3
3 +6 +4 +3 -10 6
4 /2 +4 +3 +3 10
20 3 8 10

You can check your solution to 1, 2, and 3

You can expect more of these puzzles in one of our e-books

Captain Brownbeard’s hourglasses


Brownbeards hour glasses

Captain Brownbeard of the pirateship “The northern star” told his first officer: “Tonight I want to sleep for 7 hours. Wake me exactly 7 hours after I leave the bridge.”

The unfortunate first officer knew that captain Brownbeard had only 1 punishment: eat so many chocolates that you never can stand the sight of chocolate again. He didnt want to suffer such a dreadful fate. To make things worse, he had only two hourglasses, one of 5 hours and one of 3 hours. How can he measure exactly 7 hours? Does he need to make any preparations in advance, and if so, which?

This puzzle is based on a puzzle with hourglasses of 4 and 7 hours which I found in both “More games for the superintelligent,” by James F. Fix and in “Denkspielereien” by Tom Werneck. The same problem can be found on numerous places on the web, with hourglasses of 7 and 11 minutes. This raises of course the question who invented this puzzle, and when?
Oh, by the way, the times used in this puzzle introduce a slight variation.

You can check your solution