# 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.

• What did you try? Where are you stuck? – 5xum Mar 26 at 14:46
• @5xum: I wanted to determine the position of each digit according to the title of the post which involve nine unknowns belonging to a quite restrictive set. Only the N is rather easy. – Piquito Mar 26 at 15:38

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.

• I wanted to determine the positions of the digits in the triangles, as the title says. But I put the question wrong. Anyway, you have "the" answer of the easy part and I put you an upvote. Thank you. – Piquito Mar 26 at 15:34