Q13 (a)

Q.13(a) Either: Let $a$ be divisible by $5$, i.e. $a=5k$.

$$a^2-5a$$ $$=[5k]^2-5\times5k$$ $$=25k^2-25k$$ $$=25k(k-1)$$

As the product of 2 consecutive numbers is even:

Let $k(k-1)=2m$

so $a^2-5a=25\times2m$

$$=50m$$

which is divisible by $5$ [sic].

OR

Let $a$ be divisible by $5$.

Case 1: not by 10

$a=10k-5$, $k\in\mathbb Z$.

$$a^2-5a=(10k-5)^2-5(10k-5)$$ $$=(10k-5)(10k-5-5)$$ $$=5(2k-1)\times5(2k-1)$$ $$=50(k-1)(2k-1)$$

which is divisible by $50$.

Case 2: $a$ divisible by $10$

$a=10k$.

$$a^2-5a=(10k)^2-5(10k)$$ $$=100k^2-50k$$ $$=50k(2k-1)$$

which is divisible by $50$.