ABCDEF x G=BCDEFABeing as it's the computing department a few people have brute forced this but, given that I'm ill today, I thought I'd have a go at solving it properly - and succeeded!

We know that:

ABCDEF is even

G is odd

All are different digits

Here goes....

1) For ABCDEF to be even, F must be even.

2) Even x odd = even so BCDEFA must be even, so A is even.

3) All are different digits so ABCDEF can't be equal to BCDEFA, so G does not equal 1.

4) ABCDEF has the same number of digits as BCDEFA so, A x G must be less than 10.

5) Possible options for A and G are: A ∈ {0, 2, 4, 6, 8}, G ∈ {3, 5, 7, 9}. Given that found in 4), A must be in {0, 2} as 3x4=12. Including a leading zero is rather unlikely, so A = 2.

6) Given 5) and 4), G = 3. So now we have:

2BCDEF

x 3

-----------

BCDEF2

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

7) 3 x F ends in 2, given that F is even, 3 x F ∈ {0, 12, 18, 24}. F must be 4. So now we have:

2BCDE4

x 3

-----------

BCDE42

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

8) (3 x E)+1 ends in 4, so 3 x E ends in 3. E can be either odd or even so 3 x E ∈ {0, 3, 15, 18, 21, 24, 27}, therefore E = 1.

2BCD14

x 3

-----------

BCD142

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

9) 3xD ends in 1. D can be odd or even so 3 x D ∈ {0, 15, 18, 21, 24, 27}, therefore D = 7.

2BC714

x 3

-----------

BC7142

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

10) (3 x C)+2 ends in 7, so 3 x C ends in 5. C can be either odd or even so 3 x C ∈ {0, 15, 18, 24, 27}, therefore C = 5.

2B5714

x 3

-----------

B57142

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

11) (3xB)+1 ends in 5, so 3 x B ends in 4. B can be odd or even so 3 x B ∈ {0, 18, 24, 27}, therefore B = 8.

285714

x 3

-----------

857142

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

Note, once I'd got A and G I did initially start by trying to solve this from the other side as follows (I quickly realised that I'd got the wrong end!):

7) The value of B must be at least 6 (AxG plus any carry over values). Therefore, B ∈ {6, 7, 8, 9}.

8) Given that 3 x B ∈ {18, 21, 24, 27}, some carry over value will be applied so the value of B must be greater than 6. Therefore, B ∈ {7, 8, 9} and 3 x B ∈ {21, 24, 27}.

9) Given 8), the carry over must be 2. So B is greater than or equal to 8, B ∈ {8, 9}. (Note, the obvious thing to do here is say that B ∈ 8, but it's possible for B to equal 9... take for example the value 298145 x 3 = 917435).

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