Algebra 101

Y is the sum of 3 different digits, therefore Y is one of {6, 7, 8, 9}

All green cells see r5c4. therefore, greens and reds are two different combinations and both sum to Y.

• Therefore, Y can't be 6 nor 7.

If Y is 8, then yellows sum to 16, which must be {1, 2, 3, 4, 6}. And they must be grouped as {1, 3, 4}, {2, 6}, and this collides with the 2 in r5c5.

Therefore, Y must be 9.

By Set Equivalent Theory (SET), we know that the digits of blues is equivalent to digits of yellows. Since they have the same set of digits, they must also sum to the same number.

All blue cells are covered by arrows. Therefore, we can write down the formula:

• 3 * (Y + Y + r3c5) - 2 - 2*r6c6 = 45 + 2 * x + r6c6

• Note that we have to remove the extra digits on the left hand side.

We can move some terms in the left to the right:

• 3 * (Y + Y + r3c5) = 45 + 2 * x + 2 + 3 * r6c6

Now the left hand side is a multiple of 3, so the right hand side must be a multiple of 3 too.

• Therefore, 2 * x + 2 is a multiple of 3 ==> x + 1 is a multiple of 3.

By Sudoku, 9 must be in the cage of r6c{7,8,9}, so X must be 12 or higher, and X + 1 is a multiple of 3.

• X must be 14, 17, 20, ...

r3c5 must be 8 or lower

• 78 ≥ 3*(Y + Y + r3c5) = 45 + 2 * x+2+3 * r6c6

• 31 ≥ 2 * x + 3 * r6c6

• And x ≥ 14, r6c6 ≥ 1

The inequality only holds when x = 14, r6c6 = 1, r3c5 = 8.

Now we have a grid like this, and x is 14. So we know the totals of all killer cages.

The cage in box 6 must be {2,3,9}.

r{1,2}c5 must be {1,7}, because {2,6}, {3,5} are ruled out. ==> r{1,2,3}c4 must be {2,3,4}

5 and 6 on row 6 must be in box 4.

r4c{1,2,3} can't be {1,2,6}, {1,3,5}, so it must be {2,3,4}.

r5c{1,2} must be 18.

We can write down more pencil marks.

The rest of the puzzle should be quite straight forward.