A calculus of the absurd

C.0.2 Divisibility of \(abc_{10}\)
  • Example 31 “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

\begin{equation} a + b + c = 9m \end{equation}

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

\begin{equation} abc_{10} = 100a + 10b + c \end{equation}

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 \(9\).