# A calculus of the absurd

##### C.0.2 Divisibility of $$abc_{10}$$
• Example C.0.1 “When we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If $$abc_{10}$$ is taken to represent a decimal number (not $$a$$ times $$b$$ times $$c$$ as in algebra), we mean the value $$100a + 10b + c$$. Show that if $$a+b+c$$ is divisible by 9, then $$abc_{10}$$ is divisible by 9.”

Solution: Anything that is divisible by 9 can be written as $$9 \cdot \text {something}$$. As this is algebra, we can use a letter (e.g. $$m$$) to represent “something”. As we are assuming that $$a+b+c$$ is divisible by 9, we can write that

$$a + b + c = 9m$$

Where $$m$$ is some number (which depends on the value of $$a$$, $$b$$ and $$c$$). Then we can consider $$abc_{10}$$. First, we can write this (as given in the question) as

$$abc_{10} = 100a + 10b + c$$

To show that this is divisible by $$9$$ we need to write it as $$9 \cdot \text {something}$$. We need to apply the fact that $$a+b+c=9m$$ here (because we are trying to show that, given this assumption, $$abc_{10}$$ is divisible by $$9$$). We want to somehow extract an $$a + b + c$$, here. Currently, we have a $$c$$, which is great (because it’s what we’re after), and all we need is a $$b$$ and a $$c$$, which we can get by breaking $$100a$$ into $$99a + a$$ and $$10b$$ into $$9b + b$$. When we apply this to the expression as a whole, our desired result pops out fairly quickly.

\begin{align} abc_{10} &= 100a + 10 b + c \\ &= 99 a + a + 9b + b + c \\ &= 9(11a + b) + a + b + c \\ &= 9(11a + b) + 9m \\ &= 9(11a + b + m) \end{align}

Therefore, if $$a + b + c = 9m$$ (i.e. $$a + b + c$$ is divisible by $$9$$) $$abc_{10}$$ is also divisible by $$9$$.