# Algebra Sudoku: Dozens

Alas, the grid deduction fortnightly is over, but my love for the genre remains the same - i.e. very high.
So, here is another grid deduction puzzle. An algebra sudoku!

I think the rules are fairly obvious. Just note that the numbers in each box should be read normally, going left to right or going top to bottom. Additionally, if the numbers in a box are say, '3' and '4', they can only be read as '34' - they can't be read as '3.4', '3/4', '3*4', or anything like that.

The theme of this puzzle is dozens - 12s and 13s are everywhere!

The maths in this one is very simple. But who knows how far algebra sudokus could go... who's up for a polynomial themed algebra sudoku?!? :D

• The rules aren't obvious to me... Jan 29, 2017 at 2:53
• It's just algebra :P Jan 29, 2017 at 3:12

I advise you to look at my solution with the sudoku handy otherwise you may miss some bits. Also I only do the first bit; I believe the rest is straightforward.

Firstly note that

if we have a x and 12x in the same line then x is either 6,7,8

This is trivial from a simple case bash (I'll be using case bashes a few times).

Furthermore,

If x in the above example is 6, 7 or 8, then 12x is 72, 84 or 96. In other words if a is 6, then 12a is 72 so c must be 8

Anyways we realise that a, 12a and c cover all the digits 6, 7, 8 (again from a case bash)

We can use this to deduce that

b must be 4, because it trivially can't be 1,2,3,5 or 9 (from the 13b clue) and we just eliminated 6,7,8.

Progress! Now we can evaluate

13b

e

Since we have a

1.2e, the last digit of e must be 5 (check for yourself!). Case bash the first digit - it can't be 1,2,3,4,5 or 7,8,9 so it must be a 6!

Now we have

1.2e as well, which turns out to be 78

This gives us

12c (because it must therefore be 96) which gives us c,a,d, then h.

The rest should be quite straightforward

(and it is possible to do all the algebra clues before getting to the actual sudoku)