IP1 Class Test (May 2025) Bonus Question

Consider the following proposition and proof:

Proposition:

If the sum of all digits of a two-digit positive integer is divisible by \(3\), then the number is also divisible by \(3\).

Proof:

Let the two-digit positive integer be \(x = 10a + b\), where \(a\) is the tens digit and \(b\) is the ones digit.

\(\because (a + b) \text{ is divisible by}\hspace{0.5em} 3,\)

\(\therefore \frac{(a + b)}{3} \text{ is an integer.}\)

\(\therefore x = 10a + b = 9a + (a + b),\)

\(\therefore \frac{x}{3} = \frac{9a + (a + b)}{3}\)

\(\;\,= 3a + \frac{(a + b)}{3} .\)

\(\because a, \frac{(a + b)}{3} \text{ are integers,}\)

\(\therefore 3a + \frac{(a + b)}{3} \text{ is an integer,}\hspace{0.5em} \frac{x}{3} \text{ is an integer,}\)

\(\therefore x \text{ is divisible by}\hspace{0.5em} 3. \text{ (shown)}\)

Based on the above proposition and proof, in clear workings, show that:[5]

If the sum of all digits of a four-digit positive integer is divisible by \(9\), then the number is also divisible by \(9\).[5]