Let q represent the integer length of a side of a triangle. If r represents the number of distinct values for q such that we can create a triangle with lengths q, 9, and 13, what is the value of r?
A. 5
B. 17
C. 18
D. 22
E. 29

(B) From the triangle inequality rule, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So, the three inequalities are
q + 9 > 13,
9 + 13 > q,
and q + 13 > 9.

The third equation holds for any positive value of q. So solve the first two inequalities for q, so q > 4 and 22 > q. So the value of q is 4 < q < 22.

Since the lengths of the sides of the triangle are of integer value, the lowest possible integer/side can be 5 and the highest possible integer/side can be 21 (q = 5, 6, 7, 8, 9 ..., 19, 20, or 21). So, r, the number of numbers in this set, must equal 21 – (5 – 1) = 17. The correct answer is B.
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Why don't you solve the third equation so that -4 < q < 22? (thus becoming 0 < q < 22)

Solving the third inequality doesn't give us anything new. We know that q > 0 from the original statement.

The range for q must comply with all three inequalities, NOT just some of them. The first inequality, q + 9 > 13, yields q > 4. Therefore in the solution range q must be greater than 4. You can check that q = 1, 2, 3 or 4 do not fit in the first inequality, q + 9 > 13.

The set of all the possible values of q is {5, 6, 7, ... ,20, 21}.

When we count the number of consecutive integers from m to k inclusive, we use the following formula:

k – (m – 1)
I advise you to remember it in this way and here is the explanation:

Imagine we count consecutive integers from 1 to n. It's clear that there are n such integers.
(Note, by the way, that n = n – (1 – 1), NOT n = n – 1)

If we number the elements of the set {m, m+1, ..., k – 1, k} by 1, 2, ... the last integer will be k – (m – 1), because any integer that we number an element with is (m – 1) units less than that element.

m, m + 1, m + 2, ..., k – 1, k
1, 2, 3, ... , (k – 1) – (m – 1), k – (m – 1)

Similarly, as with consecutive integers from 1 to n, the're k – (m – 1) such integers.