Proof. $10 | a$ so $a = 10k_1$ , $10 | b$ so $b = 10k_2$ , where $k_1, k_2$ are unknown $\in \mathbb{Z}$ . $$\begin{aligned} ma \pm nb &= m (10k_1) \pm n (10 k_2) \\ &= 10(m k_1 \pm n k_2) \end{aligned}$$

Note that $(m k_1 \pm n k_2) \in \mathbb{Z}$ .

$\square$

For example: $7 \not | 13$ , $7 \not | 15$ , but $7 | (13 + 15)$ .

For example: $7 \not | 15$ , $7 \not | 16$ , and $7 \not | (15 + 16)$ .

Proof. $7 | a$ so $\frac{a}{7}$ is a whole number. $7 \not | b$ so $\frac{b}{7}$ is not a whole number.

So $\frac{a}{7} + \frac{b}{7}$ is not whole. Therefore, $\frac{a + b}{7}$ is not a whole number.

So, by statement 5: $7 \not | a + b$ .

$\square$

By the fundamental theory of arithmetic, each number only has 1 pf, therefore these number cannot equal each other.

Proof. Numbers that are divisible by $5^3$ are of the form $5^3k$ .

There are two cases:

1. $k$ contains a power of 5, therefore the power of $5^3k$ will be greater than 3.
2. $k$ does not contain a power of 5, therefore the power of $5^3k$ is 3.

Both of these scenarios have too many 5s to go into numbers with a $5^2$ in their prime factorization.

$\square$