Note: technically, it was redundant to tell you that the sum of the diagonal products is 38, see the coda at the end of this file for more. Preparation: ============ It's handy to work out what the 2x2 tiles can be before starting. Thaey're easy to enumerate as one of the diagonal products must be half the sum or below, and the other one must be half the sum or above. Just enmerate the smaller one. So for 38: 1*2=2 leaves 36=4*9 or 6*6 1*3=3 leaves 35=5*7 1*6=6 leaves 32=4*8 1*8=8 leaves 30=5*6 2*3=6 leaves 32=4*8 2*4=8 leaves 30=5*6 2*5=10 leaves 28=4*7 2*7=14 leaves 24=3*8 or 4*6 2*9=18 leaves 20=4*5 3*6=18 leaves 20=4*5 And that's the 12 tiles. I just scribbled those as outer pencil markings in the top left of the grid before it started filling up. Each time one was placed, I crossed it off. The solve ========= Firstly, we deal with any crosses (which is what I am calling the marked locations where the product sum property must be satisfied) containing duplicates. If you've not done the prep-work, If there are any crosses with duplicate entries, it's easy to see that if there are duplicate digits in a product sum, they must be along the diagonal, and so would contribute 1*1, 2*2, ... 9*9 to the sum. 1 leaves 37, 2 leaves 34, 3 leaves 29, 4 leaves 22, 5 leaves 13, and over 6 goes negative. However, 6 leaves 2, and thus 1662 is the only possible cross with duplicateentries. There are 2 crosses that span box boundaries: r5c3-r6c4 and r6c6-r7c7. r5c3-r6c4 is excluded as there's a => The 1662 cross is in r6c6-c7c7 => The position of its 6s is restricted The r5c5 cross can't have 6*2 as its leading diagonal, as (38-12=26)=2*13. => It is the 1468 cross, with [48] unresolved The 2 in r7c7 is very promiscuous with regard to its diagonal partner, only 2, 6, and 8 are excluded as crosses, leaving 1 (on a 1492 cross), 3, 4, 5, 7, 9. Most possibilities can be dismissed quickly: - 2491 puts a 1 in r8c8, forcing either 1573 or 1568 into the other cross. 1573 leaves nowhere for the 6s in box 9. 1568 remains a possibility. - 2483 puts a 3 in r8c8, requiring a cross that contains a 3 and a 6, but no 2, 4, or 8. No such cross exists. - 2564 puts a 6 in r7 or c7 - 2475 puts a 5 in r8c8, requiring a cross that contains a 5 and a 6, but no 2, 4, or 7. Only 5186 satisfies that. - 2387 puts a 7 in r8c8, requiring a cross that contains a 7 and a 6, but no 2, 3, or 8. No such cross exists. - 2467 also puts a 6 in r7 or c7 - 2459 puts a 9 in r8c8, requiring a cross that contains a 9 and a 6, but no 2, 4, or 5. No such cross exists. => Box 9 can contain: 2491/1568 or 2475/5186, plus diagonal reflections. Either way, 1568 is no longer available for use elsewhere. We now turn our attention to box 4. The crosses in r5c2-r6c3 and r5c3-r6c4 may not contain a 6, and at most one may contain a 1. They must also overlap in 2 cells. If both contain a 2, they must pair along an edge with that 2. => the participants are 2483, 2475, 2387, or 2459 These can only pair up 8234:2745 or 8234:2549, either way up If one of them contains a 1 it could be the 1573 or 1492. 1573 can only pair 5317:3278, but this either puts a 1 or an 8 in c4, which box 5 forbids. If instead it's 1492, it could pair along its 42 edge or its 29 edge. The 1492:4328 pairing is rejected because of the 8 or 1, leaving 1492:4527 Therefore the r5c3-r6c3 domino is [24] in all 3 cases. Looking at the r4c2-r5c3 cross, it must share an edge with the cluster we've just made. - The 1492:4527 cluster has no symmetries available, and 14 is an unpairable edge, so rejected. => The r5c2-r6c2 domino is definitely [83] - if oriented 8234, 82 edge which can only come from 7382, and the duplicated 3 forbids that, so it must be 3482 in r5c2-r6c3. The only 34 edge available is in a 5634 so we can place that in r4c2 The r5c3-r6c4 cross remains restricted to 4527 or 4925. Remember that box 9 can only contain 2491/1568 or 2475/5186, so both box 4/5 and box 9 share an interest in 2475. We can now turning our attention to box 7, given what we know about boxes 4 and 9. 2 and 6 are both excluded from r7 and c3, so must be on the cross at r8c1-r9c2. Therefore it's 2564 or 2467 in some orientation The cross at r7c2-r8c3 that it intersects with cannot contain both a 2 and a 4, restricing it to 1573 or 2387 The 8 and 3 in c2 reject 2387 in any orientation, and the 5 prevents the intersection being 5, making 2564 impossible in r8c1-r9c2 Therefore box 7 contains 4726\1573, the latter fixed, the former with diagonal reflection possible. Box 8 is now starting to get quite restricted, the 5 cannot be in c4 or r7. Therefore the cross at r8c5-r9c6 has a 5 on it, but no 6. so is 2475 or 2459 - 2459 can only be oriented 2459 or 5294 - 2475 can only be oriented 2475 Box 3 is also restricted, as all of the remaining crosses contain a 2, so that must be the intersection. All but one remaining cross contain a 4 as well, so can't pair with any other like that. => 2387 must be one of the crosses used. => 2475 having a 7 is rejected as its partner, leaving 1492, 2564, and 2459 Let's regroup, and consider which remaining crosses can go where: 1492 can go in box 3 or 9 2564 can only go in box 3 2475 can go in box 4/5, 8, or 9 2387 must go somewhere in box 3 2459 can go in box 4/5 or 8 As 2564 must partner 2387 in box 3, we know that 1492 must go in box 9. => we can place its 2\1 diagonal. Likewise we can place 1568's 1\8 diagonal. => we can resolve the 89 pair in box 7 => we get a 456 triplet in row 9 and can place 7, 1, and 3. => 7 places in box 9. => we get a 2456 quad in row 8, so can resolve the 49 pair in box 9. => the cross in box 8 can now only be 2475 in that orientation => 2459 can only go in the box 4/5 cross, oriented 4952 => sudoku completes the bottom boxes Only now do we tackle box 3, which we know contains a 2387 and a 2564 overlapping at their 2. The 7 of the 27 diagonal can only be in r3c7, so that cross is 8273 in order to not clash on 3s. The 2564 has to be oriented 5426 to fit. 1 & 9 are placeable. At that point, one has no option than to just do sudoku. Enjoy! ======= Why I tell you the sum is 38 ============================ 38 is special, as it is the total which has the largest number of product sums - 11 - where all 4 terms are distinct. That means they can be fit entirely inside a sudoku box, not needing to cross a box boundary. Closest to that 26 and 34 have 9 such crosses, with 2 and 5 respectively that can only span a box boundary, but there are many sums that can have 8 crosses (5), or 7 (11). So techically I don't need to tell you the sum, you can easily count the 11 crosses in the grid, but working out that that corresponds to a sum of 38 is tedious. Calculating all the product sums that equal a particular target it's not too tricky for one target, but there's no rule of thumb to point to which numbers are likely as as mass producers.