The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 have in common.

First, solve for \( \left|z - 6\right| \):

\( \left|z - 6\right| \) + 4 = -5
\( \left|z - 6\right| \) = -5 - 4
\( \left|z - 6\right| \) = -9

The value inside the absolute value brackets can be either positive or negative so (z - 6) must equal - 9 or --9 for \( \left|z - 6\right| \) to equal -9:

So, z = 15 or z = -3.

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

To divide fractions, invert the second fraction and then multiply:

\( \frac{2}{9} \) ÷ \( \frac{1}{9} \) = \( \frac{2}{9} \) x \( \frac{9}{1} \)

To multiply fractions, multiply the numerators together and then multiply the denominators together:

\( \frac{2}{9} \) x \( \frac{9}{1} \) = \( \frac{2 x 9}{9 x 1} \) = \( \frac{18}{9} \) = 2

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{5!}{6!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6} \)
\( \frac{1}{6} \)