# Sum Like It Simple

The three intersecting ellipses form seven curved regions. Place one tile from the seven candidates in each of these regions so that the tiles in any ellipse adhere to the corresponding rule. Show that this solution is unique!

Mark the numbers like this:

a        c
b
d
e   f

g


The equations are:
$$\begin{eqnarray}a + b + d + e &=& 15\tag 1\\b + c + d + f &=& 19\tag 2\\d + e + f + g &=& 20\tag 3\end{eqnarray}$$

Adding $$(2)$$ and $$(3)$$ and subtracting the sum from $$a$$ to $$g$$, we get $$d + f - a = 19 + 20 - 28 = 11$$
which implies $$a \leq 2$$ because $$d + f \leq 6 + 7 = 13$$.

If $$a = 2$$, then $$d, f$$ are $$6, 7$$ and from $$(2), (3)$$ we get $$b + c = 6$$ and $$e + g = 7$$. It follows that $$b, c$$ are $$1, 5$$ and $$e, g$$ are $$3, 4$$. There is however no possible way to satisfy $$(1)$$: $$b = 1$$ is too small and $$b = 5$$ is too big.

Thus $$a = 1$$ and $$d, f$$ are $$5, 7$$ and $$(2), (3)$$ become $$b + c = 7$$ and $$e + g = 8$$. It follows that $$b, c$$ are $$3, 4$$ and $$e, g$$ are $$2, 6$$. In view of $$(1)$$, $$e = 2$$ is too small, hence $$e = 6$$ and $$b + d = 8$$. It is then clear that $$b = 3$$ and $$d = 5$$.

Therefore the only solution is: $$(a, b, c, d, e, f, g) = (1, 3, 4, 5, 6, 7, 2)$$.

• Same solution, using the fact that 39 is only 1 less then maximum possible sum of $b+c+e+g+2(d+f)$
– z100
Jun 11, 2022 at 16:00