Answer:
[tex]q = 8[/tex]
[tex]r = 3[/tex]
Step-by-step explanation:
Given
[tex]n = dq + r[/tex]
[tex]0 \le r < d[/tex]
[tex]n = 67[/tex] --- not -67
[tex]d = 8[/tex]
Required
Find q and r
Substitute values for d and b
[tex]n = dq + r[/tex]
[tex]67 = 8 * q + r[/tex]
[tex]67 = 8q + r[/tex]
Make q the subject
[tex]q = \frac{67 - r}{8}[/tex]
[tex]0 \le r < d[/tex] means that r is less than 8 but greater than or equal to 0
And r and q are integers.
Let [tex]r = 3[/tex]
[tex]q = \frac{67 - 3}{8}[/tex]
[tex]q = \frac{64}{8}[/tex]
[tex]q = 8[/tex]
No other true values of r and q can be gotten.