Respuesta :

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.