# Puzzling Place Value

*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:**

Prove that there are exactly four such numbers $AB$.

**Problem 2:**

**Problem 3:**

**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:**

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:**

**Problem 7:**

What general rule must apply? Why?

*With thanks to Don Steward, whose ideas formed the basis of this problem.*

Thank you and well done to everyone who submitted solutions to this problem. There were lots and lots of correct solutions, so we couldn't mention you all. In fact, choosing between them was difficult!

**Problem 1**

Harry and Jack, from Chew Valley School, sent this solution:

$10A+B=(A+B)\times4$, because $A$ has to be multiplied by ten to make it a 2 digit number.

$10A+B=4A+4B$, because you multiply out of the brackets and are left with this.

$10A-4A=4B-B$

So $6A=3B$

So $2A=B$

In other words, the units have to be double the tens, so the numbers are $12$, $24$, $36$ and $48$.

**Problem 2**

Tim, from Gosforth Academy, sent this solution:

$ABC + AB + C = 100A + 10B + C + 10A + B + C = 300$

$110A + 11B + 2C = 300$

A has to be below $3$, because $110 \times 3$ gives $330$, which is greater than $300$.

However, it can't be $1$, because that would leave $11B + 2C$ equalling $190$, and with the maximum value of 9, this clearly cannot be true: $(11 \times 9) + (2 \times 9) = 117$.

Therefore, $A = 2$.

Substitute this in and we get:

$(2 \times 110) + 11B + 2C = 300$

$11B + 2C =80$

The maximum value of $2C$ is $2 \times 9 = 18$, so the minimum value of $11B$ is $80 - 18 = 62$.

Of course, $11B$ must be a multiple of $11$, and it also has to be even, because $2C$ and $80$ are both even. The only even multiple of $11$ between $62$ and $80$ is $66$, which must equal 11B, so B = 6.

Substituting this into the equation gives:

$(11 \times 6) + 2C = 80$

$2C = 14$

$C = 7$

So we have only one possible set of values: $A = 2$, $B = 6$ and $C = 7$

To check this, $267 + 26 + 7$ does equal $300$.

**Problem 3**

Zainab, Amy, Ellie, from Sandbach High School, sent this solution:

Choose a two-digit number with two different digits ($AB$) and form itsreversal (i.e. $BA$).

$AB$ and $BA$ are $10A+B$ and $10B+A$

Subtracting $(A+B)$ gives $9A$ and $9B$.

Adding these gives $9A+9B=9(A+B)$, a multiple of 9.

**Problem 4**

Emma, from Sandbach High School, sent this solution:

The 6 numbers are:

$10A+B$

$10B+C$

$10C+A$

$10A+C$

$10B+A$

$10C+B$

The total is $22A+22B+22C =22(A+B+C)$. Dividing by $A+B+C$ gives $22$.

**Problem 5**

Julian, from the British School, Manila, in the Philippines, sent this solution:

Since $a>b>c$, to make the differences non-negative the differences will be:

$(10a+b)-(10b+a)$, $(10b+c)-(10c+b)$, and $(10a+c)-(10c+a)$

These are $9a-9b$, $9b-9c$, and $9a-9c$.

When you add these results you get $18a-18c=18(a-c)$ which is clearly divisible by $18$.

**Problem 6**

Zach sent us this this solution:

$100A+10B+C=(10A+B)+(10B+C)+(10C+A)$

Simplifying gives:

$100A+10B+C=11A+11B+11C$

$89A = 10C+B$

Since we can only use the digits $1$ to $9$, the only possible solution for $A$ is $A=1$, because any larger $A$ gives a $3$ digit number.

This leaves $C=8$ and $B=9$.

To check, $19+98+81=198$.

**Problem 7**

Sergio, from the King's College of Alicante, Spain, sent this solution:

Firstly I transformed it into an equation: $10a+b+10c+d=10d+c+10b+a$.

This simplifies to $9a-9b+9c-9d=0$, and dividing by $9$ gives $a+c=b+d$.

This will always work because the first equation is the same as the last one, just in a simplified form, and you could undo all of the steps.

It is like the quadratic equation, $ax^2+bx+c$ is the same as the quadratic formula, just that it ´s rearranged to make $x$ the subject. Well here it ´s the same but putting each letter only once.

Why do this problem?

This problem helps students to appreciate the power of algebra for solving problems involving number and place value. It would make a good follow-up problem for students who have worked on Reversals.

Possible approach

*This printable worksheet may be useful: Puzzling Place Value*

You may wish to use this problem for consolidation in class or homework, after working on Reversals or Always a Multiple.

When students have had a go at tackling the questions, take some time to discuss the answers as a class, focussing particularly on their explanations of why there were only a limited number of solutions, or why the solutions they have found satisfied a particular condition.

### Key questions

Is there an algebraic expression to represent the problem?

Are there any restrictions on A, B and C?

### Possible support

Always a Multiple offers multiple representations for solving place value problems, so it might be a good problem to try before looking at this one.

Possible extension

For students who are ready to explore algebraic representations with quadratic expressions, Factorising with Multilink and Pair Products may be suitable follow-up problems.