The Wrong Products

Question

Rules:

1. Place each number from 1 to 12 exactly once such that every row and column contains exactly two numbers.
2. Clues outside the grid show the product of the two numbers in that row/column. However, the clues are wrong. The right products are either one less or one more than the clues.

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The Puzzle:

Answer 1

(I know Vaekor answered first, however providing some logic behind the solution - I was 95% done before seeing their answer)

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Starting off by deducing some of the correct products:

The 97 must be 96, as 98 cannot be made, and 96 must be 12 x 8
The 109 now must be 110, as the 12 has been used in the columns, and 110 is 11 x 10
The 20 must be 21 as 19 is prime, and 21 is 7 x 3
The 64 must be 63, as 65 cannot be made, and 63 is 7 x 9

There is now only 1-6 available for the columns, and we can deduce that 11 should be 10, which is 5 x 2, 13 should be 12, which is 3 x 4 and 5 should then be 6 which is 6 x 1

We have now fully deduced the columns, and have an overlap in the rows, so we can start filling in numbers and deducing from there:

51 must be 50 which is 5 x 10, as 52 has no valid factor pairs.
10 is either 9 or 11, both of which require a 1
71 is now 72 rather than 70, as 10 is used, so we have 6 and 12



34 has no valid factor pairs, so 35 must be 36, and that now must be 4 and 9
The 9 is now used, so the 10 is 11 and 11 x 1
And finally, only 2 and 8 are left so 15 must be 16

And we have the final answer!

Answer 2

I believe the solution should be:

+--+-+-+-+-+--+
|  |9| |4| |  |
+--+-+-+-+-+--+
|  | | | |6|12|
+--+-+-+-+-+--+
|  |7| |3| |  |
+--+-+-+-+-+--+
|  | |2| | | 8|
+--+-+-+-+-+--+
|10| |5| | |  |
+--+-+-+-+-+--+
|11| | | |1|  |
+--+-+-+-+-+--+

Answer 3

Here is my solution by integer linear programming.

Let binary variable $b_{ijk}$ represent whether number $k$ is placed in cell $(i,j)$, while $k=0$ means leaving the cell empty. And let binary variable $rs_{i}$ represent whether the true product of row $i$ is the shown number $+1$ and $cs_{j}$ represent whether the true product of column $j$ is the shown number $+1$.

MIP does not like products, but by taking log, it converts to sums. So let constants $c_k = \log k$ for $k=1,\dots, 12$, $c_0 = 0$, and $\alpha^c_j=(\log(p^c_j +1) - \log(p^c_j - 1)$, $\beta^c_j = \log(p^c_j - 1)$, where $p^c_j$ means the given product of column $j$, and set $\alpha^r_i, \beta_r^i$ accordingly. Then $(\alpha^c_j cs_j + \beta^c_j)$ and $(\alpha^r_i rs_i + \beta^r_i)$ mean log of the true product depending on $cs_j$ and $rs_i$.

The constraints are $$ \sum_{ij} b_{ijk} = 1 \quad \forall k=1,\dots,12\\ \sum_{k} b_{ijk} = 1\qquad \forall i,j\\ \sum_{i} b_{ij0} = 4\qquad \forall j\\ \sum_{j} b_{ij0} = 4\qquad \forall i\\ \left|\sum_{i,k} c_{k} b_{ijk} -(\alpha^c_j cs_j + \beta^c_j)\right| \leq \varepsilon\quad \forall j \\ \left|\sum_{j,k} c_{k} b_{ijk} -(\alpha^r_i rs_i + \beta^r_i)\right| \leq \varepsilon\quad \forall i$$

The solution given by other answers is the only solution according to the solver.

PS. I accidentally forgot the constraint $\sum_{ij} b_{ijk} = 1$ at first, but there is no other solution generated, meaning that it is redundant to rule every digit exactly once.