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**01/01/2013 - Tents (*)**

TEE = 4 (Therefore T = 2, E = 1)

TEN = 6 (Therefor N = 3)

NEST = 10 (Therefore S = 4)

and

**(2 + 1 + 3 + 2 + 4)**

__TENTS = 12__----------------------------------------

**02/01/2013 - Triangles (**)**

There are a few variations of this solution, but they all require the moving of the same

**. One variation is shown below.**

__3 tokens__The moving of these three tokens will result in the forming of the triangle shown below.

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**03/01/2013 - Rope Around The World (**)**

This solution is dependent on you knowing that the circumference of a circle is 2πr (where r is the radius). In this instance we know that 2πr = 40,000km (or 40,000,000 metres).

We want to know what the circumference would be if 2πr was such that our radius is increased by 1.

Mathematically we can write our question as:

2π(r + 1) = ? metres

or aftter expanding the brackets:

2πr + 2π = ? metres

We know that 2πr = 40,000,000 metres in this case, therefore:

40,000,000 + 2π = 40,000,000 + ? metres

cancelling to give:

2π = ? metres

π = 3.14, therefore:

**6.28 metres!**

Strange as it seems, we only need 6.28 (or 2π) extra material. And this is true of any size or circle, if we wanted to raise the rope by 1 metre.

----------------------------------------

**04/01/2013 - Magic Square (**)**

Below is one of a number of solutions, and I will explain how I created this solution quickly and logically.

Firstly, we have the numbers 1 to 9. The average of these numbers is 5. Therefore 5 HAS to be in the centre, with the numbers either side (in any direction) summing to 10.

The number 9 cannot be placed in a corner, because the 1 would have to go in the opposite corner (9 + 5 + 1 = 15) and there would have to be two addition ways of creating 6 (apart from 1 + 5, which has already been used) and there is only 1 (2 + 4). Therefore, the 9 goes on a side (not in a corner) and the 1 opposite it.

We have just proved that to go with the 9 there is only one set of numbers left (2 and 4), so I put the 2 next to the 9 and now it is imply a case of filling in the blanks.

- 2 + 9 + ? = 15, gives ? = 4
- 2 + 5 + ? = 15, gives ? = 8
- 4 + ? + 8 = 15, gives ? = 3
- 4 + 5 + ? = 15, gives ? = 6
- 2 + ? + 6 = 15, gives ? = 7

A quick check and my solution was complete.

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**05/01/2013 - Connected Circles (***)**

A bit of logic required for this one.

I started by realising the 19 was the total of all the numbers except 2, so this had to be at the bottom.

Secondly that gave the 3 it's position.

And then by noticing the symmetry on the two out numbers (both being 10), meaning that the 3 plus another number must equal the total of symmetrically opposite numbers, which means only 3 + 6 = 4 + 5. So the 6 and 1 had places to go and now the 4 and 5 were easy to place.

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**06/01/2013 - Catch the Thief (***)**

I have a couple of things I want to say here. The first is well done to anyone who got this correct and managed to collect

**all 23 coins**. The second is to say thank you to my friend Tom, who found the answer shown below.

When I looked at this puzzle, I was convinced the answer would be 22 and was pleased to find that there was a better solution. Yours may look slightly different to the one above, but hopefully some aspects are the same.

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**07/01/2013 - Bug Jump (***)**

The solution above might not be the quickest but it seems to be one of the most logical. Feel free to post your own versions.

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**08/01/2013 - Even Chocolates (**)**

The easiest solution to find (like the one above) will involve symmetry, but there are multiple solutions and they do not all need to be symmetrical.

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The energetic gardener would be able to mow the lawns on his own in 12 days.

In 1 day he could mow 1/12th of the lawns, and in 8 days 8/12ths.

Therefore, the lazy gardener most mow 4/12ths in 8 days.

To complete all the lawns on his own (12/12ths) it will therefore take 3 times 8 days...

The answer below is the actual one from The Canterbury Tales, where each square is visited each square once and once only in 15 moves.

----------------------------------------

Below are my answers, although there are a number of solutions:

0 = 4/4 - 4/4

1 = 4/4 x 4/4

2 = 4/4 + 4/4

3 = SQRT(4 x 4) - 4/4

4 = 4! - 4 x 4 - 4

5 = SQRT(4 x 4) + 4/4

6 = (4 + 4)/4 + 4

7 = 4!/(4 + 4) + 4

8 = (4 + 4) x 4/4

9 = (4 + 4) + 4/4

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Hopefully this one wasn't too difficult.

Here is the easiest solution to visualise:

----------------------------------------

There are the 64 (1 by 1) squares you'd expect on an 8 by 8 board. But there are also 49 (2 by 2) squares and 36 (3 by 3) squares and the square number pattern continues to give the following solution:

64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 =

For example, if there are 2 baskets with apples in (4oz apples in one basket and 5oz apples in the other) we can take 1 apple from one basket, 2 from the other. By weighing the apple we will have either a weight totaling 13oz (2 x 4oz apples and 1 x 5oz apple - meaning the 1st basket has the 5oz apples in) or 15oz (2 x 5oz apples and 1 x 4oz apple - meaning the 2nd basket has the 5oz apples in).

Extending this idea further, we can take a different number of apples from each basket, 1 from the first basket, 2 from the second etc, up to 10 from the last. In total there are 55 apples weighed. If all the apples weighed 4oz the total weight would be 220oz. For every 1oz over that total we know the basket number, e.g. 226oz total weight equals the basket you took 6 apples from, to weigh.

If you put all the dart numbers in order you can simply workout the moving average, starting with aiming at the number 1 and adding the numbers either of it, before dividing by 3 gives 13 through 20 + 1 + 18 = 39, 39/3 = 13. Carrying on this moving average you get 1 + 18 + 4 = 23, 23/3 = 7.67; etc.

The correct answer is to

----------------------------------------

We can solve this puzzle using algebra and simultaneous equations.

Let J = Number of Jars and P = Number of Pencils.

(1) J = (P/9) + 2

(2) P = 6J + 3

(1) x 9:

9J = P + 18 = (3)

Substitute (3) into equation (2):

9J = (6J + 3) + 18

Simplify to get:

3J = 21

J = 7

Substitute (J = 7) into (2):

P = 6 x 7 + 3

P = 45

Therefore,

----------------------------------------

----------------------------------------

There are probably a few good way to do this, but what I did was to make the base of the triangle equal zero by using the numbers -1, 1, -2 and 2 - putting the -1 and 1 at the ends so that they can couple with the -3 and 3. This means that the -4 and 4 fall into place above them and leave the 0 at the top.

It makes sense to have the 0 on a corner as there is no negative equivalent of this number.

My solution is shown below.

----------------------------------------

**09/01/2013 - Gardeners (**)**The energetic gardener would be able to mow the lawns on his own in 12 days.

In 1 day he could mow 1/12th of the lawns, and in 8 days 8/12ths.

Therefore, the lazy gardener most mow 4/12ths in 8 days.

To complete all the lawns on his own (12/12ths) it will therefore take 3 times 8 days...

**24 days!****----------------------------------------**

**10/01/2013 - Pardoner's Tale (***)**The answer below is the actual one from The Canterbury Tales, where each square is visited each square once and once only in 15 moves.

----------------------------------------

**11/01/2013 - Four 4s (1-10) (**)**Below are my answers, although there are a number of solutions:

0 = 4/4 - 4/4

1 = 4/4 x 4/4

2 = 4/4 + 4/4

3 = SQRT(4 x 4) - 4/4

4 = 4! - 4 x 4 - 4

5 = SQRT(4 x 4) + 4/4

6 = (4 + 4)/4 + 4

7 = 4!/(4 + 4) + 4

8 = (4 + 4) x 4/4

9 = (4 + 4) + 4/4

----------------------------------------

**12/01/2013 - Die Hard Jugs (*)**Hopefully this one wasn't too difficult.

Here is the easiest solution to visualise:

- Fill the 3 litre jug.
- Pour the contents of the 3 liter bottle into the 5 litre jug.
- Fill the 3 litre jug, again.
- Fill the remainder of the 5 jug from the 3 litre jug, leaving 1 litre in the 3 litre jug.
- Empty the 5 litre jug.
- Pour the contents (1 litre) of the 3 litre jug into the 5 litre jug.
- Fill the 3 litre jug.
- Empty the contents of the 3 litre jug in the 5 litre jug, leaving 4 litres in the 5 litre jug.

Here is the quickest solution:

- Fill the 5 litre jug.
- Pour the contents of the 5 litre jug into the 3 litre jug, leaving 2 litres in the 5 litre jug.
- Empty the 3 litre jug.
- Pour the contents (2 litres) from the 5 litre jug into the 3 litre jug.
- Fill the 5 litre jug.
- Fill the 3 litre jug with the contents from the 5 litre jug, leaving 4 litres in the 5 litre jug.

----------------------------------------

**13/01/2013 - Maths and Arrows (**)**
Using logic we can build the solution up.

Firstly, looking at the "40" target.

If we scored 40 with an arrow we have 60 left to score, which we can work out from a bit of basic numeracy isn't possible using the numbers 16, 17, 23, 24, 39 and 40.

The same logic rules "39" out, too.

So, we are left with 16, 17, 23 and 24. "24" cannot be placed with "16", because this would leave "40" which we know leaves an impossible number to get an answer from. The same logic stops "23" and "17" being placed together.

Further examination stops "23" being placed with "16 and "24" or placing "24 with 17".

Now, we know that the only numbers are "16" and "17". The least number of numbers to reach at least 100 is going to use the "17".

1 x 17 = 17

2 x 17 = 34

3 x 17 = 51

4 x 17 = 68

5 x 17 = 85

6 x 17 = 102

- 17 + 16 = - 1

Therefore, the answer must be

So, we are left with 16, 17, 23 and 24. "24" cannot be placed with "16", because this would leave "40" which we know leaves an impossible number to get an answer from. The same logic stops "23" and "17" being placed together.

Further examination stops "23" being placed with "16 and "24" or placing "24 with 17".

Now, we know that the only numbers are "16" and "17". The least number of numbers to reach at least 100 is going to use the "17".

1 x 17 = 17

2 x 17 = 34

3 x 17 = 51

4 x 17 = 68

5 x 17 = 85

6 x 17 = 102

- 17 + 16 = - 1

Therefore, the answer must be

**17 + 17 + 17 + 17 + 16 + 16 = 100 in 6 arrows**.----------------------------------------

**14/01/2013 - Chess Board (**)**There are the 64 (1 by 1) squares you'd expect on an 8 by 8 board. But there are also 49 (2 by 2) squares and 36 (3 by 3) squares and the square number pattern continues to give the following solution:

64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 =

**204**.**----------------------------------------**

**15/01/2013 - Spoke (*)**
The key to this puzzle is to find the average of the the numbers (in this case 3.5) and place numbers summing to twice that (in this case 2 x 3.5 = 7) on either side of the central 7.

The solution is shown below.

**----------------------------------------**

**16/01/2013 - Heavy Apples (***)****Understanding that there are different numbers of apples in each basket allows us to logically find a solution.**

For example, if there are 2 baskets with apples in (4oz apples in one basket and 5oz apples in the other) we can take 1 apple from one basket, 2 from the other. By weighing the apple we will have either a weight totaling 13oz (2 x 4oz apples and 1 x 5oz apple - meaning the 1st basket has the 5oz apples in) or 15oz (2 x 5oz apples and 1 x 4oz apple - meaning the 2nd basket has the 5oz apples in).

Extending this idea further, we can take a different number of apples from each basket, 1 from the first basket, 2 from the second etc, up to 10 from the last. In total there are 55 apples weighed. If all the apples weighed 4oz the total weight would be 220oz. For every 1oz over that total we know the basket number, e.g. 226oz total weight equals the basket you took 6 apples from, to weigh.

**----------------------------------------**

**17/01/2013 - Average Darts (**)**If you put all the dart numbers in order you can simply workout the moving average, starting with aiming at the number 1 and adding the numbers either of it, before dividing by 3 gives 13 through 20 + 1 + 18 = 39, 39/3 = 13. Carrying on this moving average you get 1 + 18 + 4 = 23, 23/3 = 7.67; etc.

The correct answer is to

**aim for the number 1**.----------------------------------------

**18/01/2013 - Pencils in a Jar (***)**We can solve this puzzle using algebra and simultaneous equations.

Let J = Number of Jars and P = Number of Pencils.

(1) J = (P/9) + 2

(2) P = 6J + 3

(1) x 9:

9J = P + 18 = (3)

Substitute (3) into equation (2):

9J = (6J + 3) + 18

Simplify to get:

3J = 21

J = 7

Substitute (J = 7) into (2):

P = 6 x 7 + 3

P = 45

Therefore,

**7 Jars and 9 Pencils**.----------------------------------------

**19/01/2013 - Numbers 1 (**)****5 x 4 + 7 x 6 - 7 = 62****5 ÷ 4 x (7 + 6 + 7) = 25****5 - 4 x 7 + 6 x 7 = 19****20/01/2013 - Flipping Coins (**)**

**The minimum number of coin flips is 4.****The solution I used is as follows (labeling the coins 1 to 4 from left to right):**

- Flip coins 1, 2 and 3 leaving Tail (T), Tail (T), Tail (T) and Head (H).
- Flip coins 2, 3 and 4 leaving T, H, H, T
- Flip coins 1, 2 and 4 leaving H, T, H, H
- Flip coins 1, 3 and 4 leaving T, T, T, T

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Using algebra we can write our missing value as "x".

And 100,000 + x (putting 1 at the front of a 5 digit number) is three times bigger than putting the 1 at the end of the 5 digit number - which can be written as 10x + 1.

Therefore 3(100000 + x) = 10x +1.

And now we can solve the equation:

10x + 1 = 3(100000 + x)

10x + 1 = 300000 + 3x

10x = 299999 + 3x

7x = 299999

x = 299999/7

x = 42857

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Below is a series of diagrams showing my answer.

----------------------------------------

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The quickest route is to utilise Amie and Ben as much as possible (as torch carriers).

----------------------------------------**21/01/2013 - 5 Digit Number (***)**Using algebra we can write our missing value as "x".

And 100,000 + x (putting 1 at the front of a 5 digit number) is three times bigger than putting the 1 at the end of the 5 digit number - which can be written as 10x + 1.

Therefore 3(100000 + x) = 10x +1.

And now we can solve the equation:

10x + 1 = 3(100000 + x)

10x + 1 = 300000 + 3x

10x = 299999 + 3x

7x = 299999

x = 299999/7

x = 42857

**The answer is 42857.**----------------------------------------

**22/01/2013 - 5 Digit Number (***)**

**5 + 7 x 9 + 5 + 6 = 79****(5 + 7) x 9 - (5 x 6)****= 78****5 7 9 5 6****= 48**

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**23/01/2013 - Switcharoo (**)**Below is a series of diagrams showing my answer.

----------------------------------------

The easiest way to solve this is to list the square numbers:

1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

4 x 4 = 16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81

10 x 10 = 100

And now look to see if any 3 of these numbers add to make 180, which they do.

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It clear that just adding doesn't work so there must be some multiplication involved. If we are multiplying then 8 x 3 = 24 so that is ruled out and 3 x 3 = 9 and that doesn't help. From this we can infer fractions are likely to be used. And knowing that 8 x 3 equals 24 we also know that 8/(1/3) equals 24 and here is our elegant solution.

3 - 8/3 = 1/3

8/(1/3) =24

Written in one:

**24/01/2013 - Square Triangle (*)**The easiest way to solve this is to list the square numbers:

1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

4 x 4 = 16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81

10 x 10 = 100

And now look to see if any 3 of these numbers add to make 180, which they do.

**The answer is 16, 64 and 100.**

----------------------------------------

**25/01/2013 - 24 from 8, 8, 3 and 3 (**)**It clear that just adding doesn't work so there must be some multiplication involved. If we are multiplying then 8 x 3 = 24 so that is ruled out and 3 x 3 = 9 and that doesn't help. From this we can infer fractions are likely to be used. And knowing that 8 x 3 equals 24 we also know that 8/(1/3) equals 24 and here is our elegant solution.

3 - 8/3 = 1/3

8/(1/3) =24

Written in one:

**8/(3-(8/3)) = 24**----------------------------------------

**26/01/2013 - Connected Circles 2 (***)**

Once you have the 1 and 2 the correct way around for the top-right of this puzzle, the other numbers indicate that 3 and 5 can't be placed in the circles summing to 14 (as all the numbers are equal) and they must be in the centre. This helps complete the solution rather quickly.

----------------------------------------

**27/01/2013 - Pool Balls (**)**
There are
4 possibilities for this experiment

Second
ball chosen is red and the first red is picked

Second
ball chosen is red and the second red is picked

Second
ball chosen is yellow and the red is picked

Second
ball chosen is yellow and the yellow is picked

As
the 4th
possibility cannot be true because its given the first ball picked is red, then
we have 3 equal possibilities with which 2 give red as the remaining ball, so

**= 2/3****----------------------------------------**

**28/01/2013 - Bridge Walk (**)**The quickest route is to utilise Amie and Ben as much as possible (as torch carriers).

- AB Cross first (2mins)
- B comes back (2mins)
- CD cross (10mins)
- A comes back (1min)
- AB cross together (2mins)

**Total = 17mins****29/01/2013 - Guess My Number (**)****This is a logically puzzle, which can be taken in the steps given and listing the possible numbers**

- I am a three digit number = Any 3 digit number (???)
- I am divisible by 5 = ?? (last digit 0 or 5)
- My digits add up to 8 = 125, 170, 215, 260, 305, 350, 440 or 530
- The tens digit is smaller than hundreds digit = 215, 305 or 530
- I am less than 300
**= 215**.

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There are many ways of doing this - one is show below.

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**30/01/2013 - Four in a Row (**)**

**The answer is 10.**There are many ways of doing this - one is show below.

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**31/01/2013 - Zero (**)**There are probably a few good way to do this, but what I did was to make the base of the triangle equal zero by using the numbers -1, 1, -2 and 2 - putting the -1 and 1 at the ends so that they can couple with the -3 and 3. This means that the -4 and 4 fall into place above them and leave the 0 at the top.

It makes sense to have the 0 on a corner as there is no negative equivalent of this number.

My solution is shown below.

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