<s>[INST] Finding place of the nine digits The nine digits 1, 2, 3, ... .., 9 are placed in the nine triangles of the attached figure in such a way that the digits around each circle add up as indicated. Calculate the value of N.
 
 [/INST] The total of the digits from $1$ to $9$ is $45$.  Four digits on the left side of the diagram add to $16$; four on the right add to $25$.  That leaves $N = 45 - 16 - 25 = 4$.
EDIT: Let's find the rest.
With $4$ assigned, the additional four digits in the total $32$ must add up to $28$. The only possibility is $5,6,8,9$; correspondingly, the four digits not adjacent to $32$ are $1,2,3,7$.
The total of $25$ is made up of two digits each from these two sets of four.  No choice without $7$ is large enough; considering whether $1,2$ or $3$ is included quickly reveals two possibilities: $25 = 1+7+8+9$ or $3+7+6+9$.  These correspond to $16 = 2+3+5+6$ and $1+2+5+8$.
The total of $20$ comes from four digits: two from $1,2,3,7$ and two from $5,6,8,9$, but also two each from the quadruples making up $25$ and $16$.  If the correct way of forming $25$ and $16$ is $(1+7)+(8+9)$ and $(2+3)+(5+6)$, then $20$ must be made with one digit chosen from each of the four parenthesized groups.  None of the sixteen possibilities work.  So it must be $25 = (3+7)+(6+9)$ and $16 = (1+2)+(5+8)$.  Only one possibility works: $20 = 7+6+2+5$.
Thus the unique solution has the digits $2,5,6,7$ from left to right in the top half and $1,8,4,9,3$ in the bottom half.
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