Fiendishly difficult puzzles – how did you do?
This article is more than 7 years oldIn his celebration of the rich history of puzzles, Alex Bellos set readers some classic brainteasers. If you haven’t finished with them then read no further, because the answers are below ...
See the puzzles here
1 QWERTY
TYPEWRITER. The question asked for a word that used only the 10 keys, not that uses every one of them. There are other answers, but those ones don’t appear in the question.
2 Business card
June, July, August, September, October, November, December, January, February, March, April, May.
3 The double date
The two sons are both uncle and nephew to each other. For a “simple” puzzle it is surprisingly brain-muddling! Let’s call the two men Albert and Bernard, and their sons Steve and Trevor. I’ve drawn the family tree below. Bernard and Steve are half-brothers because they have the same mother. Bernard’s son Trevor is therefore Steve’s nephew.
Likewise, Albert and Trevor are half-brothers, so Steve is Trevor’s nephew.
The confusing family ties get even more twisted when we consider that Bernard’s mum is married to Albert, so Bernard’s mum’s stepmother is Albert’s mother. This makes Albert’s mother Bernard’s grandmother. Bernard is therefore married to his grandmother, and is therefore grandfather to himself!
4 From Le Havre to New York
When I first heard this puzzle I instantly thought that the answer was seven, since the crossing takes seven days, which was surely also the response of the French mathematicians who first heard this puzzle from Monsieur Lucas.
Wrong! What about the liners that left New York in the past week? They are currently at sea and you will pass all of them on your journey too. The correct answer is that as you leave Le Havre you pass one in the port (that’s the one arriving at noon, having left New York a week ago); you will pass 13 at sea; and you will meet a final one that is departing as you reach New York at noon in a week’s time.
Here’s a diagram that explains it:
If the liners are all travelling at constant speed, you will pass a liner every 12 hours, once at noon (Le Havre time) and once at midnight.
5 Smith, Jones and Robinson
The most straightforward way to solve the puzzle is to draw two grids. We need to find which of Smith, Jones and Robinson are the driver, fireman and guard, so draw one grid, as shown below left, that contains the names of the workers and the professions. The question also involves three passengers and three locations. So draw a second grid, shown below right, that has Mr Smith, Mr Jones and Mr Robinson, against Leeds, Sheffield and halfway between.
Our first piece of solid information is that Mr Robinson lives in Leeds, so we can tick the Mr Robinson/Leeds box, and put crosses in the boxes that have Mr Robinson living somewhere else, or anyone else living in Leeds. We need to take other clues together to fill in more squares. For example, the guard’s nearest neighbour, who is a passenger, earns exactly three times as much as the guard does. So we can eliminate Mr Jones as the guard’s nearest neighbour, because his salary is indivisible by three.
But neither can the guard’s nearest neighbour be Mr Robinson, because the guard lives between Leeds and Sheffield and Mr Robinson is in Leeds. So the guard’s nearest neighbour, and fellow resident of “between Leeds and Sheffield”, must be Mr Smith. We can put a tick in the top right of the right-hand box, as shown below, and deduce therefore that Mr Jones is in Sheffield because that is the only option remaining.
The guard’s namesake lives in Sheffield. We know that Mr Jones lives in Sheffield. So the guard must be Jones. We can tick Jones/guard, as shown below left, and cross off the other boxes in the same row and column since Jones has no other occupation and the others are not the guard.
The clue that Smith beats the fireman at billiards reveals that Smith is not the fireman. (Robinson must be the fireman). So we can cross off Smith/fireman. We already know that Smith is not the guard. So by elimination Smith is the engine driver.
6 Free the coin
Using a second match, light the head of the match and quickly blow it out. It will stick to the glass on the right, enabling you to remove the glass on the left and take the coin.
7 Eight neat sheets
The sheet directly underneath 1 can only be the sheet in the top left corner. And so on as the sheets spiral anticlockwise:
8 The number tree
Start at the top left. How can 72 and 99 make 27?
Got it! 99 – 72 = 27. In other words, the number in a circle is the difference of the numbers in the two circles that point to it. And this works down the line with 45 –27 = 18, and 39 – 18 = 21, which means that the missing number must be 15, until you get to 21 – 13 which is equal to 8 not 7!
Back to square one. Or rather circle one. How else can 72 and 99 make 27? The answer is so simple it is easy to miss. 7 + 2 + 9 + 9 = 27. The digits all add up. Likewise, 2 + 7 + 4 + 5 = 18, and so the missing number must be 2 + 1 + 3 + 6 = 12.
The puzzle is fantastically ingenious because Yoshigahara has found two arithmetical rules that fit the same numbers for five steps in the sequence, but only one of which fails for the final step, and then only by 1.
Can You Solve My Problems? A Casebook of Ingenious, Perplexing and Totally Satisfying Puzzles by Alex Bellos is published by Guardian Faber at £14.99. To order a copy for £10.99 go to bookshop.theguardian.com, or call 0330 333 6846.
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