Digit3 can only be 3, 6 or 9 (as it's 3 x Digit2).

Digit2 can only be 1, 2 or 3.

Digit1 can only be 2, 4 or 6.

But, since Digit4 is two thirds of Digit1, Digit1 must be 6. Making Digit4 4 and making Digit3 9. Making Digit2 3.

Giving 6394. QED.

Puzzle 2

If you were to spell out the numbers in full, (One, Two, Three, etc), how far would you have to go until you found the letter 'A'?

[Ref: ZHVF]

Hint: Start counting now.

Answer: 1000.

One thousAnd.

Unless you use British English when the answer becomes 101 = one hundred And one.

Puzzle 3

My local bus company has recently expanded and no longer has enough room for all of its buses.

Twelve of their buses have to be stored outside.

If they decide to increase their garage space by 40%, this will give them enough room for all of their current buses, plus enough room to store another twelve in the future.

Hint: A tricky puzzle that will need a little algebra (or clever thinking).

Answer: 72 buses.

They have enough room for 60 of these, expanding the 60 capacity by 40% will give them enough room for 84, which we know is 12 more spaces than they currently need.

If they have B buses and S spaces before the expansion, they have enough room for:

S = B - 12 [1]

After the expansion they have more spaces, and enough room for:

S + 0.4 x S = B + 12 [2]

Rewriting [2] as:

1.4S = B + 12

and again as

B = 1.4S - 12 [3]

We can rewrite [1] as:

B = S + 12 [4]

Making [3] = [4] we have:

1.4S - 12 = S + 12

Subtracting S from both sides gives:

0.4S - 12 = 12

Adding 12 to both sides gives:

0.4S = 24

Multiplying by 10 and dividing by 4 on both sides gives:

S = 60

Using S = 60 in [4] gives B = 72. QED.

Puzzle 4

As my birthday approaches I start to collect leaves - a little bizarre perhaps, but I enjoy it!

On the first day of the month I collect 1 leaf, on the second day I collect 2 leaves, the third day I collect 3 leaves, and so on.

By my birthday I will have collected 276 leaves altogether. On which day of the month is my birthday?