February 2013 Solutions

Click on the images to see a larger version.

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01/02/2013 - Spoke 2 (**)



Nothing too difficult with this one, once you realise that the combines sum of two opposite spokes must equal the combined sum of any other opposite spokes.

Given 7, it means that the opposite spoke must be 1 and that only 6 and 2 can also equal 8, same with 5 and 3, leaving 4 to go in the middle of the spokes.

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02/02/2013 - It Equals 1 (***)


This is tough to answer and really just takes a lot of trial and error.

9/12 + 5/34 + 7/68 = 1

This is easier to see if you put all the fractions over 204.

153/204 + 30/204 + 21/204 = 204/204.

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03/02/2013 - Fifteen (*)

There are 10 ways, which can be found logically by starting with the lowest number and working through the permutations.
  1. 069
  2. 078
  3. 159
  4. 168
  5. 249
  6. 258
  7. 267
  8. 348
  9. 357
  10. 456
Note, that 456 = 465 and 654 and 645 and 564 and 546 because they all have the same digits used.  If you did work them all out separately then you'd have the answer 60!

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04/02/2013 - Got Any Change? (**)

There's a bit of logical thinking required here, but starting with the largest coins we know we can only have 1 fifty pence piece.  The same logic proves we can only have 4 twenty pence pieces.  We can't have any 10 pence pieces, because 50, 20, 20 and 10 would equal 100 pence.  However we can have a five pence piece and 4 two pence pieces.

Therefore the solution is you can have £1.43 from 50p +  4 x 20p + 5p + 4 x 2p.

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05/02/2013 - From A to B (***)


Call the downhill travelled x, flat straight y and the uphill z. 

4 hours 40mins = 14/3 hours
  • On the way there we get the equation x/20 + y/15 + z/12 = 4
  • On the way back we get x/12 + y/15 + z/20 = 14/3
Quickest way is to notice 12 = 3 x 4, 15 = 3 x 5 and 20 = 4 x 5 so we can rewrite these as…
  • 3x + 4y + 5z = 4 x (3 x 4 x 5)
  • 5x + 4y + 3z = (14/3) x (3 x 4 x 5)
Add these to get 8(x +y +z) = (26/3) x 3 x 4 x 5

x + y + z = 13 x 5 = 65miles

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06/02/2013 - How Many Triangles (**)

16 x single triangles
7 x double triangles
3 x triple triangles
1 x quadruple triangle

27 triangles in total.

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07/02/2013 - A Dozen Apples(***)


Recognise that 12/x = y

So our equation afterwards becomes 12/(x - 1) = y + 1

Substituting y = 12/x into the equation we get a quadratic. 

Rearrange to get x^2 – x – 12 = 0

We get roots x = -3 or x = 4

So solution is x = 4, i.e. 12 apples originally cost £4, and you could get 3 apples for £1

The new way allows you to buy 12 apples for £3 and get 4 apples for £1.

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08/02/2013 - Dominoes (*)

You can play around with this quite easily and there are a few solutions.  Mine is shown below.


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09/02/2013 - Trains (**)

Number the trains 1 to 6, with 1 being at the back and 6 at the front. Then take the clues one at a time.

Clue 1: The red engine is behind the yellow engine.
  • So the Red engine can only be 1, 2, 3, 4 or 5
  • And the Yellow engine can only be 2, 3, 4, 5 or 6
Clue 2: The yellow engine is behind the green engine.
  • So the Yellow engine can only be 2, 3, 4 or 5
  • Therefore the Red engine can only be 1, 2, 3 or 4 (from our first clue).
  • And the Green engine can only be 3, 4, 5 or 6
Clue 3: The Orange Train is last.
  • So the Red engine can only be 2, 3 or 4
  • The Yellow engine can only be 3, 4 or 5
  • The Green engine can only be 4 or 5
Clue 4: The blue engine is between the purple engine and the red engine.
  • If the Red engine is behind the Yellow engine which is behind the Green engine, then the Blue engine must be behind the Red engine and in front of the Purple engine.
  • The Red engine can only be 4
  • The Yellow engine can only be 5
  • The Green engine can only be 6
  • The Blue engine can only be 3
  • The Purple engine can only be 2 
A diagram is shown below.



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10/02/2013 - Even Triangle (*)

There's a bit of logic required here.  You need to spot that you can't have the two largest (12 and 13) numbers on the same row or have the two smallest numbers (8 and 9) on the same row.

The easiest way to solve this is to put the 8 at the top and the 9 in the centre at the bottom.

I'm now left with 10. 11, 12 and 13.  Meaning my 13 and 10 must go on one side and 11 and 12 on the other side.  Further more the 13 can't go next to the 9 because it would be counted with both the 8 and 9 and there's no way to even that up... therefore the 13 must go below the 8 to one side.

From here the numbers are easy to place.  A diagram shows my solution below.


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11/02/2013 - Number Circle (*)

The easiest way of solving this is to start with the number 3, because we know the only numbers that can go next to this are 1 and 5, so we can put these in straight away.

The next step is to look at the number 1 and note that this can only go next to either 3, 4 or 5.  Because 3 and 5 are already used it must be next to 4.

This eave the number 2 to go in the last remaining place.

A quick check and it all works nicely.


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12/02/2013 - Carriages (**)

There are a number of possible solutions and here is how I worked mine out.

I started with the last clue "Mr White is not by the door".


And since no man can be in the centre (or he will be next to or opposite his wife - because the men cannot sit next to each other) he must be at one end.


Opposite him could be a man or a woman, but if Mr Green sits there then Mrs White must be next to him.


Mr and Mrs Browns' positions are now given and Mrs Green fits in nicely.


My solution is show below.




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13/02/2013 - Wolves (*)

This is a proportion problem.

If six wolves can kill six lambs in six minutes, then one wolf can kill lamb each minute, therefore to kill sixty lambs in sixty minutes requires sixty wolves.

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14/02/2013 - Valentines (*)

This is just a straight bit of algebra, rather than a puzzle, but it's nice to solve.
  • 9x + 7i < 4x + 21u +5x     - collect the like terms to give:
  • 9x + 7i < 9x +21u     - the 9x cancels on each side leaving:
  • 7i < 21u     - divide each side by 7 leaves:
  • i < 3u    - or:
  • i <3 U     - (I heart you).
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15/02/2013 - Birthday Cake (**)

To solve this problem think of the cake as a 3 dimensional object.

Cut the cake in half from the top, before slicing again at a right-angle to make 4 equal parts (quarters).  Finally, turn the cake on it's side and slice through the middle (i.e. where the filling is) and you are left with 8 equal parts.

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16/02/2013 - When? (*)

This can be answered by taking it step-by-step.

Three days ago, yesterday was the day before Sunday, so three days ago was itself Sunday.


That means today is Wednesday, so tomorrow is Thursday.


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17/02/2013 - 53 Steps (**)

A bit a trial-and-improvement is a good method here.  Plot a path, work out the total and change your path slightly to adjust the number.


Above is an example of the solution I found.


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18/02/2013 - Missing Number (**)

The answer is shown below, take a look and see if you can work out why before reading on.


The answer is 5 because we add the bottom two numbers (those along the triangles base) and divide the total by the number at the top (the other corner).


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19/02/2013 - How Many Triangles 2 (**)


There are 12 single triangles, 6 made from 4 individual triangles and 2 made from 9 individual triangles, giving a total of 20 triangles.


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20/02/2013 - Toads and Toadstools (***)

This needs a bit of playing around, but there is a pattern to the shortest path, which is 16 moves:
(1—5), (3—7, 7—1), (8—4, 4—3, 3—7), (6—2, 2—8, 8—4, 4—3), (5—6, 6—2, 2—8), (1—5, 5—6), (7—1)

Using the number layout below.



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21/02/2013 - Golden Payment (**)

If you can only cut the Gold Bar into 3 pieces you cut it into 1/7, 2/7 and leave 4/7.  This way:
Day 1: Give the 1/7 piece. (You have 6/7, they have 1/7)
Day 2: Give the 2/7 piece and take the 1/7 piece back. (You have 5/7, they have 2/7)
Day 3: Give the 1/7 piece. (You have 4/7, they have 3/7)
Day 4: Give the 4/7 piece and take back the 1/7 and 2/7 pieces. (You have 3/7, they have 4/7)
Day 5: Give the 1/7 piece. (You have 2/7, they have 5/7)
Day 6: Give the 2/7 piece and take the 1/7 piece back. (You have 1/7, they have 6/7).
Day 7: Give the 1/7 piece. (You have 0/7, they have 7/7).

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22/02/2013 - Sharing Sandwiches (***)

The simplest way to solve this puzzle is to not think about 5 and 3 sandwiches but how many sandwich parts after all the sandwiches have been cut into 3 pieces.

So, Fred brought 5 sandwiches or 15 sandwich parts and Bill brought 3 sandwiches or 9 sandwich parts.

Because each person is getting an even amount of sandwich parts (8 each), Fred has to donate 7 sandwich parts and Bill only 1.

Therefore, Fred should get £7 and Bill £1.


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23/02/2013 - Bad Shop (*)

This can be solved in simple steps.

Georgia is buying something for $70 but hands over $200, so Georgia is currently $130 down and the attendant $130 up.

However, the attendant hands over $150.  So, Georgia ends $20 up and the attendant $20 down.


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24/02/2013 - Missing Numbers (**)

The easier way to work this puzzle out is to write the question in column format.

 835X
+XY2=
Y14Y

Written like this it is possible to test different number, starting with X.

E.g. If, X =1 and X+2=Y then Y=3: however 5+3 does not equal 4. Therefore X is not 1.

Repeating this process you eventually find that X=7 and Y=9.

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