In the diagram, each of the integers 1 through 9 is to be placed in one circle so that the integers in every straight row of three joined circles add to 18 . The 6 and 1 have been filled in. The value of the number represented by $x$ is
(A) 4
(B) 5
(C) 7
(D) 8
(E) 3

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-053.jpg?height=306&width=482&top_left_y=671&top_left_x=1250)
Consider the possible ways that the numbers from 1 to 9 can be used three at a time to sum to 18 .

Since we may not repeat any digit in the sum, the possibilities are:

$1+8+9,2+7+9,3+6+9,3+7+8,4+5+9,4+6+8,5+6+7$.

Next, consider the row $1+d+f$ in Fig. 1 .

Since the sum of every row is $18, d+f=17$.

This gives that either $d=8$ and $f=9$ or $d=9$ and $f=8$.

Next, consider the 3 rows in which $x$ appears.

The sums of these 3 rows are $a+x+d, b+x+f$, and $c+x+6$.

That is, $x$ appears in exactly 3 distinct sums.

Searching our list of possible sums above, we observe that only the numbers 6,7 and 8 appear in 3 distinct sums.

That is, $x$ must equal either 6,7 or 8 .

However, the number 6 already appears in the table.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-252.jpg?height=299&width=461&top_left_y=978&top_left_x=1491)

Fig. 1

Thus, $x$ is not 6 .

Similarly, we already concluded that either $d$ or $f$ must equal 8 .

Thus, $x$ is not 8 . Therefore, $x=7$ is the only possibility.

Fig. 2 shows the completed table and verifies that the number represented by $x$ is indeed 7 .

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-252.jpg?height=293&width=478&top_left_y=1377&top_left_x=1474)

Fig. 2