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Puzzling Place Value

Age 14 to 16 Challenge Level:
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.

Problem 1:

A positive 2 digit integer $AB$ is four times the sum of its digits.
Prove that there are exactly four such numbers $AB$.

Problem 2:

Show that there is only one set of values $A, B$ and $C$ that satisfies $$ABC + AB + C = 300$$

Problem 3:

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 $9$.

Problem 4:

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 $22$.

Problem 5:

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 $18$.

Problem 6:

For any three digits $A, B$ and $C$, show that there is only set of values that satisfies $$ABC=AB+BC+CA.$$

Problem 7:

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.