Given the list $2,3,4,5$, there are exactly three different ways to choose three integers from the list and form a triangle whose side lengths are equal to those integers. The integers chosen could be $2,3,4$ or $2,4,5$ or $3,4,5$. The integers $2,3,5$ cannot be used as side lengths of a triangle. Given the list $4,10,3, n, 13$, there are exactly four different ways to choose three integers from the list and form a triangle whose side lengths are equal to those integers. If $n$ is different from all other numbers in the list, then the sum of all possible values of $n$ is
(A) 46
(B) 29
(C) 69
(D) 23
(E) 17
For any triangle, the sum of the lengths of two sides is always greater than the length of the third side. This property is known as the triangle inequality.

If for example the side lengths of a triangle are $a, b$ and $c$, then the triangle inequality says that

$$
a+b>c \text { and } a+c>b \text { and } b+c>a
$$

We begin by considering the number of different ways to choose three integers from $3,4,10,13$ (without using $n$ ), and then forming a triangle whose side lengths are equal to those integers. Consider choosing the integers $3,4,10$.

Since $3+413$ and $4+13>10$ and $10+13>4$.

Thus, without using the value of $n$, there is exactly one way to choose three integers and form a triangle whose side lengths are equal to those integers.

This means that we need to determine values of $n$ for which there are exactly three different ways to choose two of the integers $3,4,10,13$ and form a triangle whose side lengths are equal to those two integers and $n$.

There are six possible ways to choose two integers from the list $3,4,10,13$.

Thus for each value of $n$, the triangles we need to consider have side lengths: $3,4, n$ or $3,10, n$ or $3,13, n$ or $4,10, n$ or $4,13, n$ or $10,13, n$.

For each value of $n$, we need exactly three of these six triangles to satisfy the triangle inequality. Next, we use the triangle inequality to determine the restrictions on $n$ for each of the six possible groups of triangles.

In a triangle with side lengths $3,4, n$, we get $3+n>4$ or $n>1$, and $3+4>n$ or $n3$ or $n>-1$.

To satisfy all three inequalities, $n$ must be greater than 1 and less than 7 .

Thus, the possible values of $n$ for which a triangle has side lengths $3,4, n$ are $n=2,3,4,5,6$.

Since $n$ must be different from all other numbers in the list, then $n=2,5,6$.

In a triangle with side lengths $3,10, n$, we get $3+n>10$ or $n>7$, and $3+10>n$ or $n3$ or $n>-7$.

To satisfy all three inequalities, $n$ must be greater than 7 and less than 13 .

Thus, the possible values of $n$ for which a triangle has side lengths $3,10, n$ are $n=8,9,11,12$ $(n \neq 10$ since 10 is in the list).

In a triangle with side lengths $3,13, n$, we get $3+n>13$ or $n>10$, and $3+13>n$ or $n3$ or $n>-10$.

To satisfy all three inequalities, $n$ must be greater than 10 and less than 16 .

Thus, the possible values of $n$ for which a triangle has side lengths $3,13, n$ are $n=11,12,14,15$ $(n \neq 13$ since 13 is in the list).

In a triangle with side lengths $4,10, n$, we get $4+n>10$ or $n>6$, and $4+10>n$ or $n4$ or $n>-6$.

To satisfy all three inequalities, $n$ must be greater than 6 and less than 14 .

Thus, the possible values of $n$ for which a triangle has side lengths $4,10, n$ are $n=7,8,9,11,12$.

In a triangle with side lengths $4,13, n$, we get $4+n>13$ or $n>9$, and $4+13>n$ or $n4$ or $n>-9$.

To satisfy all three inequalities, $n$ must be greater than 9 and less than 17 .

Thus, the possible values of $n$ for which a triangle has side lengths $4,13, n$ are $n=11,12,14,15,16$.

In a triangle with side lengths $10,13, n$, we get $10+n>13$ or $n>3$, and $10+13>n$ or $n10$ or $n>-3$.

To satisfy all three inequalities, $n$ must be greater than 3 and less than 23 .

Thus, the possible values of $n$ for which a triangle has side lengths $10,13, n$ are $n=5,6,7,8,9,11,12,14,15,16,17,18,19,20,21,22$.

Recall that for each value of $n$, we need exactly three of the six triangles to satisfy the triangle inequality (the triangle with side lengths $4,10,13$ is the fourth).

Clearly for values of $n$ less than 7, there are too few triangles, and similarly for values of $n$ greater than 16, there are also too few triangles. (There are at most two triangles in each of these two cases.)

In the table below, we summarize our work by placing a checkmark if the triangle satisfies the triangle inequality and then counting the number of such triangles.

| $n$ | $(3,4, n)$ | $(3,10, n)$ | $(3,13, n)$ | $(4,10, n)$ | $(4,13, n)$ | $(10,13, n)$ | $(4,10,13)$ | number of  triangles |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 7 |  |  |  | $\checkmark$ |  | $\checkmark$ | $\checkmark$ | 3 |
| 8 |  | $\checkmark$ |  | $\checkmark$ |  | $\checkmark$ | $\checkmark$ | 4 |
| 9 |  | $\checkmark$ |  | $\checkmark$ |  | $\checkmark$ | $\checkmark$ | 4 |
| 11 |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | 6 |
| 12 |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | 6 |
| 14 |  |  | $\checkmark$ |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | 4 |
| 15 |  |  | $\checkmark$ |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | 4 |
| 16 |  |  |  |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | 3 |

Therefore, there are exactly four different values of $n$ that satisfy the given conditions, and the sum of these values of $n$ is $8+9+14+15=46$.

ANSWER: (A)