You may wish to look at Always a Multiple and Reversals before tackling this problem.
For these problems, use the digits from $1$ to $9$
We are using $ABC$ to represent a 3 digit number with $A$ hundreds, $B$ tens and $C$ units. Similarly, $AB$ represents a 2 digit number with $A$ tens and $B$ units.
A positive 2 digit integer $AB$ is four times the sum of its digits.
Prove that there are exactly four such numbers $AB$.
Show that there is only one set of values $A, B$ and $C$ that satisfies $$ABC + AB + C = 300$$
Choose a two-digit number with two different digits (e.g. $47$) and form its reversal (i.e. 74). Now, subtract the sum of the digits from each of these numbers, and then add the two results. Show that you always obtain a multiple of $9$.
Choose three different digits and form the six two-digit numbers that use two of the three digits. Add these six possibilities and divide this total by the sum of the three digits. Show that you always obtain $22$.
Choose three different digits $A, B$ and $C$, with $A > B > C$ (e.g. $8, 6$ and $3$).
Work out the differences between the two-digit numbers you can make and their reverses (e.g. $86-68; 63-36; 83-38$), then add these three results.
Show that you always obtain a multiple of $18$.
For any three digits $A, B$ and $C$, show that there is only set of values that satisfies $$ABC=AB+BC+CA.$$
Find examples that fit the rule: $$AB+CD=DC+BA,$$ without any of the four digits being the same (e.g. $97+24=42+79$).
What general rule must apply? Why?
With thanks to Don Steward, whose ideas formed the basis of this problem.