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In the following sum the digits 0 to 9 have all been used, O = Odd, E = Even, zero is even and the top row's digits add to 9. Can you determine each digit?

Hint: The largest possible numbers could start with 6 and 8, therefore the first digit of the answer is 1.

Answer:

Remembering that:

E + E = E
O + O = E
E + O = O

To discuss individual letters it's easiest to represent the sum as:

A B C
D E F +
--------
G H I J

The largest values for A and D are 6 and 8, which makes G = 1.

Since column 2 is E + O = O there can be no carry from column 1 (since E + O + 1 is always even). Therefore C and F are 3 and 5 (but we don't yet know which is which), therefore J = 8.

There can't be a carry from column 2 (as A + D is even) therefore E can't be 9 as this would force a carry.

Therefore I = 9. Hence B can't be 0. Therefore H = 0.

The last remaining odd number makes E = 7. Making B = 2.

Therefore A and D are 4 and 6 (but we don't yet know which is which).

Since the top row's digits have to add to 9 the top number must be 423.

This makes the sum 423 + 675 = 1098. QED.

Puzzle 2

Four clowns were getting ready for the evening's performance. From the clothes trunk each clown had to get their socks and shoes.

There was one pair of each shoes and socks in each colour.

The red socks went with the blue shoes.

Baba had yellow shoes but not yellow socks.

The clown with green shoes did not have blue socks.

Boba had neither red shoes nor red socks.

Bilpo's shoes were the same colour as Babil's socks.

The clown with red shoes had green socks.

Can you tell who had which colour shoes and socks?

Answer:
Baba had yellow shoes and blue socks.
Boba had green shoes and yellow socks.
Bilpo had red shoes and green socks.
Babil had blue shoes and red socks.

Puzzle 3

Using all of the letters of the alphabet, once only, complete these words:

Hint: Try breaking each cog size into prime factors.

Answer: 30 revolutions.

If we break each wheel into its prime factors, we get:

63 = 3 x 3 x 7
42 = 2 x 3 x 7
35 = 5 x 7
27 = 3 x 3 x 3

We now think of rotating the large wheel just once, and this is 3 x 3 x 7 teeth moved (3 x 21), and we can see that 42 tooth wheel also has a 3 x 7 (21 teeth) in it, with an extra 2. If we therefore rotate the 63 toothed wheel twice, the 42 will have rotated three times.

The answer involves cancelling any common factors from the large wheel. We can cancel 3, 3, 7 from any of the smaller ones to leave 2 (from the 42), 5 (from the 35) and 3 (from the 27). 2 x 5 x 3 = 30. QED.

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