**Step by step analytical solution 9x9 “very dif” 2012-05-15.** The tuesday’s 9x9 is usually the most difficult calcudoku during the week. They come with different shapes requiring different techniques. This is the 9x9 difficult (tuesday May 15, 2012), Puzzle id: 461715, solving rate 132.4. The difficulty level is theorically a little higher than the equivalent 9x9 on tuesday Apr 17, 2012 (solved step by step in a different thread). Our purpose here is to use

**only the analysis** (no “guessing” or many “trial and error” will be necessary, everything will come straight forward) (in this post I will mainly use conventional language) (graphic 1 shows the problem):

First (graphic 2) we write (in blue colour) all known numbers and candidates; in steps 1 and 2

**additionally we will place some 5’s, 7’s and 9's**.

Step 1.

In row 4, a 5 can only go to f4, consequently d4 + e4 = 16, then d4 = 9 (due to e7 = 9), e4 = 7.

The cage “64x” (in two lines) has initially three possibilities: [1,1,8,8], [1,2,4,8] and [2,2,4,4] but due to the L-shape the [1,1,8,8] and [2,2,4,4] are not valid then “64x” = [1,2,4,8].

Step 2.

(In violet colour) Row 5: the 7 cann’t go inside any of the cages “11+” (due to i5 = 4) so now the only valid place is d5 = 7 with e5 + f5 = 10 >>> [1,9], since [2,8], [3,7] and [4,6] are not possible now and, consequently, e5 = 1, f5 = 9 (due again to e7 = 9).

Also we observe that since 7 is not a divisor of “162x” or “432x” and b5 <> 7, the 7 of column b must go to b9 and after this the 5 of column b must go to b5 >>> c5 = 6 (obviously the 5 is not a divisor of those multiplication cages). The cage “11+” (g5-h5) = [3,8].

The 3-cell cage “144x”, initially with these four combinations: [2,8,9], [3,6,8], [4,4,9] and [4,6,6], has been left to [3,6,8] since the 9’s are not allowed (already in use in columns d, e and f) and the 6 is being repeated in [4,6,6]. Since the 3 of column e cann’t go to the cage “2-” (due to e3 = 5 and e5 = 1) it must be e6 = 3 and e1-e2 = [6,8]. Also d6-f6 = [6,8].

Step 3.

We will now obtain the possible combinations for the other products (just in case we need it, the “addition value”, that is, the sum of each combination, is shown in parentheses):

“640x” = [1,2,5,8,8] (24), [1,4,4,5,8] (22), [2,2,4,5,8] (21);

“2268x” = [1,4,7,9,9] (30), [1,6,6,7,9] (29), [2,2,7,9,9] (29), [2,3,6,7,9] (27), [3,3,4,7,9] (26), [3,3,6,6,7] (25);

“162x” = [1,2,9,9] (21), [1,3,6,9] (19), [2,3,3,9] (17);

“432x” = [1,6,8,9] (24), [2,3,8,9] (22), [2,4,6,9] (21), [3,3,6,8] (20), [3,4,4,9] (20), [3,4,6,6] (19);

“24x” = [1,3,8] (12), [1,4,6] (11), [2,2,6] (10), [2,3,4] (9).

Step 4.

Let’s reduce a little bit these combinations. For the cage “432x”, the [3,3,6,8] and the [3,4,6,6] are inmediately cancelled (due to the 3 and the 6 in “144x” in such a way that a 3 or a 6 can not be repeated). The other four combinations of “432x” have a 9.

For the cage “162x” the [1,2,9,9] is not valid (d4 = 9) but the other two have a 9 which being in b2 or b3 force the required 9 of cage “432x” to be in c6 (we write it in green colour, graphic 3). Now the cage “12+” = [5,7] and then the 7 of cage “2268x” goes to g2.

The combination [1,2,5,8,8] is not valid for “640x” due to the double 8 (there is an 8 in “2-”) but the other two combinations for this cage also contain an 8. In these conditions there are four 8’s in the four top rows (considering the cages “64x” and “16x”) so “2:” cann’t be [1,2], [2,4] or [4,8] so it must be [3,6] and we write those candidates (in green).

Step 5.

Now let’s see row 3: The only place for a 7 in this row is c3 (in brown colour) and then d2-d3 = [1,3]. Now, since the pairs 1-3, 1-4 or 3-4 are in column d they cann’t go “simultaneously” to the cell c7 so the only valid combination for “24x” is [2,2,6] and we write those numbers. As a consequence d6 = 8 and f6 = 6.

Step 6 (in red colour).

If, for instance, we name “a” the cell g3 (for simplicity) and “b” the cell g7, applying the property of the addition to column f we find:

**a** +

**b** + 45 (45 is the full column f) = 7 + 15 + 5 + 9 + 6 + 14 + 3 = 59 >>>

**a** +

**b** = 14. This sum can not be obtained with the pair 6-8 because we already have three 8’s in columns g, h and i (in “64x”, “16x” and “11+”) so it must be

**[5,9]** with g3 = 9 and g7 = 5. As a consequence f2-f3 = [2,4] and f7-f8 = [1,8].

And we can also determine the cage “2268x” applying the addition rule to the three rightmost columns:

“2268x” (+) + 9 + 15 + 10 + 11 + 4 + 20 + 12 + 5 + 22 = “2268x” (+) + 108 = 135 (3 x 45, that is, three columns). Then “2268x” (+) = 27 >>> “2268x” = [2,3,6,7,9].

Step 7 (graphic 4).

Let’s continue: i6 = 5, i7 = 7 (in light blue colour). Also, since “162x” contains a 9, b2 = 9 (due to d4 = 9 and g3 = 9). It is clear that [2,3,3,9] is not valid now for the cage “162x” (two 3’s, but there is already a 3 in “2:”) so “162x” = [1,3,6,9] (19).

(In purple colour). The 6 in b3-b4 and the 6 in c5 cancel the combinations [1,6,8,9] and [2,4,6,9] for “432x” and since [3,4,4,9] is impossible (a double 4 would go to b6-b7-b8), then “432x” = [2,3,8,9] (22).

The 3’s in “2:” and “162x” force d3 = 1 and d2 = 3. Now the 1 in d3 makes b4 or c4 to be a 1 and, consequently, the 1 of cage “64x” goes to h2.

Now the 2’s in “64x” and “16x” force f3 = 4 and f2 = 2. This 2 makes the combination [2,2,4,5,8] impossible for “640x” >>> “640x” = [1,4,4,5,8] (22) (the only left combination for this cage).

Applying the addition rule to the three leftmost columns:

22 + 19 + 7 + 9 + 2 + 11 + “3-” (+) + 22 + 2 + 36 = “3-” (+) + 130 = 135 (3 x 45, that is, three columns). Then “3-” (+) = 5 >>> “3-” = [1,4].

Let’s continue with graphic 5.

Step 8 (in green colour).

The 1-4 in a6-a7 produce b1 = 4, c2 = 4, c1 = 1; additionally a2 = 5 and a1 = 8. And then e1 = 6, e2 = 8 and we place those numbers; e1 = 6 forces the 6 of cage “2268x” to i2.

We write the rest of the numbers in green: b9 = 7 >>> a8 = 7, a9 = 9; c7 = 2 and d8 = 2 >>> b6 = 2 and b7-b8 = [3,8]; c8 + c9 = 36 - 7 - 9 - 7 = 13 = [5,8] ([4,9] and [6,7] not valid now in column c), but also the 3 of “162x” must go to c4 (due to the 3 in b7 or b8), then again the only left numbers in column c are the pair 5-8. Besides this we can write the candidates (among 2, 3 and 9) for the cells g1, h1 and i1.

Step 9 (in brown colour).

To place the final numbers in the case of the cages “2:” and “162x” is elementary. Also since b7-b8 = [3,8] and f7-f8 = [1,8] >>> c9 = 8 and c8 = 5. We complete the cells g9, h9 and i9: g7 = 5 and i6 = 5 >>> h9 = 5; i2 = 6 >>> g9 = 6 and then i9 = 1.

Graphic 6.

Step 10 (in red colour).

To finish the puzzle is very easy considering that g8 + i8 = 10 (the rest for the cage “22+”), so g8-i8 = [1,9] due to [2,8], [3,7] and [4,6] not valid for row 8. So, g8 = 1 and i8 = 9.

Now g8 = 1 >>> f8 = 8, f7 = 1. Also f8 = 8 >>> b8 = 3, b7 = 8 and f7 = 1 >>> a7 = 4, a6 = 1. From this we deduce the candidates for the cells g6-h6, that is, [4,7], and, since g2 = 7, we have g6 = 4 and h6 = 7. Additionally i8 = 9 >>> i1 = 3 >>> g1 = 2 >>> h1 = 9.

Step 11 (in blue colour). In row 8 we need a 6: h8 = 6; in row 7 we need a 3: h7 = 3. This 3 implies g5 = 3, h5 = 8; now the 8 of cage “64x” goes to g4 and the rest is self explanatory.

And the official solution (graphic 7):