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Interlocked Rings shown in the picture share a common sum of 28.

The numbers to be filled vary from 1 to 12 and do not repeat.

Already four numbers..8, 9, 10, 11 are placed.

Fill the X s with rest of numbers such that they all add upto 28 around each ring.

enter image description here

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  • $\begingroup$ this resembles the Olympics logo... better title? $\endgroup$ – Omega Krypton Jun 18 at 11:32
  • $\begingroup$ I liked x in the box rhyme..that’s why I went for it.. $\endgroup$ – Uvc Jun 18 at 11:34
  • $\begingroup$ ok I respect your decision... it is yours anyway :) $\endgroup$ – Omega Krypton Jun 18 at 11:34
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Solution

enter image description here

Reasoning

The first thing we look at is where to place the $12$.
Notice that the circles $A$, $B$ and $D$ have totals of $8$, $9$ and $9$, respectively to be filled. This means that a $12$ can only appear in one of the squares on the extreme right of the picture.
If it is in the top of these, then the bottom one has to be $10$ to compensate (not allowed since $10$ is already filled in). Similarly, if it is in the bottom of these then the top one has to be $14$ (out of range). Hence, we are forced into positioning the $12$ into the middle of these rightmost squares.

From there, the entries in the other squares of circle $E$ must add up to $6$. This gives essentially four options, being the placement of either the pair $1,5$ or $2,4$ with optional switching.
Putting $1$ in the upper square implies the top of circle $B$ must contain $8$, not allowed.
Putting $4$ in the upper square implies the right of circle $C$ must also contain $4$, not allowed.
Putting $2$ in the upper square means the remaining squares in $B$, $C$ and $E$ contain values $7$, $6$ and $4$, respectively, leaving the numbers $1,3$ and $5$ to fill in the rest of the grid, which is undoable.

Hence, the only remaining option is to put $5$ in the upper square of $E$ and from there, the rest of the grid is forced.

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  • $\begingroup$ Yes, and it's the only solution. $\endgroup$ – Gareth McCaughan Jun 18 at 11:47
  • $\begingroup$ @GarethMcCaughan Yes, just working on that now. $\endgroup$ – hexomino Jun 18 at 11:49
  • $\begingroup$ Got it!!....... $\endgroup$ – Uvc Jun 18 at 11:51
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Disclaimer: I'm typing this out as I'm thinking this through, so it's a bit disorganized. I'll go back and work on cleaning this up when I get the time.

Initially, looking at the rings:

The total of all the rings added together is 140, doublecounting the two numbers on the left and right between AD and CE, and triplecounting the three numbers in the middle. Since all the numbers add to 78, and we know of 9 and 10 in the middle, the two numbers on the sides + 2 times the remaining middle number must equal 24. There are only three combinations of numbers that fulfill this, with the first number corresponding to the one in the middle: (7,6,4), (6,5,7), (5,2,12), and (3,6,12).

Further testing of these combinations thins this down to:

(7,6,4) and (5,2,12), as if 6 is the center number, 7 and 5 have to be the remaining numbers in Ring E and the middle rows, which is impossible, and if 3 is the center number then regardless of if the number on the right is 6 or 12, then the remaining number in E will have to be 3 or 9, both of which are already used.

Looking at the rings:

We know that the two remaining numbers in B and in D must add up to 9. The combinations with remaining numbers that sum to this are (4,5), (3,6) and (2,7).

Likewise,

The remaining numbers in A must sum to 8, being either (3,5), (2,6), or (1,7). Since the number on the left in the third row must be 6, 4, 2 or 12, we can conclude that it must be 2 and 6, which means the two numbers in ring B must be 4 and 5, since a number from each combination is present in ring A. Since the center number can only be 5 or 4, we know the top row, center number must be 4 and the second row, third number is 5.

Now working from Ring E:

The number on the third row on the right must be either 12 or 2 from the prior calculation of the middle five numbers. If the number is 2, the remaining number in E would be 11, which is already used, so the number third row on the right must be 12 and the second number on the bottom row must be 1, and now the remaining number in Ring C must be 3.

Finally:

We are remaining with the numbers 6, 2, and 7. The only combination that fulfills Ring D from before is (2,7) and since Ring A must contain 2 and 6, we can conclude that the first number in the top row is 6, the first number in the third row is 2, and the first number in the bottom row is 7.

Therefore we come to the final solution of:

(6,4,8),(11,9,5,3),(2,10,12),(7,1)

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