4
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Picture shown below contains 3 horizontal and 3 vertical relations.

Digits to be filled vary from 1 to 9.

Order of operations is given following the arrows vertically and horizontally.

enter image description here

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3
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Here is the solution

enter image description here

Reasoning

If we look at the first column we notice that the bottom entry must be a divisor of 4. It cannot be 4 since that would imply that the two digits above it are the same (not permitted) so it must be 1 or 2.

If it is 1, then the digits in the first column are fully determined as being (8,2,1). From there, the first row is also fully determined as being (8,9,3). However, once we have done this, the smallest entry we have left for the middle of the final column is 4 which makes the first product at least 12 which means we would need to fill in the bottom right-hand corner with the digit 9 and this has already been used. Hence, there is no solution in this case.

Hence, the bottom left-hand corner must be 2 and there are two possibilities for the first column: either (8,4,2) or (6,3,2) which we'll label as cases (i) and (ii).

In case (i), the first row is either (8,9,3) or (8,7,1).
If it is (8,9,3), then we already know the middle of the last column must be less than 4 (to avoid the conflict from before) but this just leaves 1 and implies the bottom right-hand corner is 0 (not allowed)
If it is (8,7,1) then the middle column must contain two other numbers with a difference of 5 but in the remaining set of numbers available, no two digits have a difference of 5. Hence, there are no solutions in case (i).

In case (ii), it must be that the other two digits have a difference of 6 which implies that the middle row is (3,7,1) as this is the only possibility from the digits remaining. From there the other digits in the last column must differ by 3 and the only possibility is to have 8 at the top and 5 at the bottom. Looking at the first and third rows, it follows that 4 is the middle of the first row and 9 is the middle of the third row, which is a valid solution.

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  • $\begingroup$ Got it !!!..... $\endgroup$ – Uvc Jun 21 at 10:37

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