Answer:
see attached
Explanation:
Given a partially-filled circle pattern, you want to place the remaining numbers so the sum in each circle is 21.
Approach
There are seven (7) circles, and nine (9) unknown values. This means there are a couple of "free choice" values that cannot be found directly. Our approach is to limit the number of possibilities we must examine.
Equations
We note that if we multiply the sum (21) by the number of circles, that product includes the "single cover" circle shapes once, and each of the overlapping "double cover" areas twice. There are three unknown "single cover" values, and six unknown "double cover" values. If we represent the sums of these by "s" and "d" respectively then we have ...
(10 +11 +6 +4 +s) +2(d +5) = 7×21
41 +d +2s = 147 . . . simplify
d +2s = 106 . . . . . . subtract 41
The total of all of the unplaced numbers is ...
(2 +12 +7 +14 +1 +8 +9 +3 +13) = s +d
69 = s +d
Now, we have two equations in s and d:
Subtracting the second from the first gives ...
(d +2s) -(d +s) = (106) -(69)
s = 32
The sum s is the sum of 'a', 'b', and 'c' as marked in the attachment. There is only one set of unplaced numbers that has a total of 32: {7, 12, 13}.
Using these numbers means that the set of allowed values for the "double cover" areas is {1, 2, 3, 8, 9, 14}.
Trial and error
Working clockwise from the space marked 'c' in the attachment, the "double cover" oblong spaces must have the values ...
- 16 -c
- c -5
- 15 -c
- 6 -(a -c)
- 9 +(a -c)
- 8 -(a -c)
As we noted above, {a, b, c} is some permutation of {7, 12, 13}. This means the value (a-c) will be one of ±1, ±5, ±6.
If (a-c) = ±6, then 6 -(a-c) = {0, 12}. Neither value is allowed: 0 is not on the list, and 12 is already used.
If (a-c) = ±5, then we would have ...
for (a-c) = -5, 6 -(a-c) = 11, not an allowed value (already used).
for (a-c) = 5, all of the values 6-5=1, 9+5=14, and 8-5=3 are allowed. This case corresponds to {a, b, c} = {12, 13, 7}.
If (a-c) = ±1, then we would have 6 -(±1) = {5, 7}. Neither value is allowed.
Conclusion
The only allowed choice for {a, b. c} is {12, 13, 7}. Using c=7 and (a-c)=5, the "double cover" values clockwise from 'c' are ...
- 16-7 = 9
- 7 -5 = 2
- 15 -7 = 8
- 6 -5 = 1
- 9 +5 = 14
- 8 -5 = 3
When all of the values are filled in the diagram it looks like the attachment.
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Additional comment
As you might imagine, there was a bit of thrashing around before this method of solution presented itself. The system of 7 equations in 9 unknowns cannot be solved directly, and we know of no way to include the requirement that the solution set must include each number of the given set, used only once.
Fortunately, the method we tried here proved to find a solution in fairly straightforward fashion.