Gran, how old are you?

When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?

Age

7 to 11

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Mum and her four children live with Gran at 13 Drywater Street.

One day, Charlie, who is the third child, asked, "Gran, how old are you?"

Gran answered, "My Grandmother would have said 'As old as my tongue and a little older than my teeth!' but I will tell you how to work out my age."

"If you multiply Mum's age with your age and with the ages of your brother and sisters you will get the answer $111 111$.

If you add Mum's age along with the ages of all you four children the total will be my age."

Charlie worked this out very quickly, because he knew his Mum's age, his age and the ages of his brother and sisters.

"Oh Gran!" he called as he ran off to play outside, "You are old!"

How old was his Gran?

We received lot of solutions to this problem, so it wasn't easy to choose from so many. Zak and Thanade from UWCSEA suggested that the first solution is to make the mom $111,111$ years old and the children $1$, $1$, $1$ and $1$. They did go on to find the 'real' solution, though!

Henry from Boasley Cross used trial and improvement very effectively:

We chose mum to be $44$ first, then $30$ and thought that she would be in between $30$ and $44$. We then tried lots of different combinations until I hit $111,888$ with the combination $37$, $7$, $12$, $12$ and $3$ and tried to lower it down but it went too low. I was quite stuck until I thought that even numbers wouldn't fit into $111, 111$ so I changed both $12$s to $11$ and $13$. When I tried the combination of $37$, $7$, $11$, $13$ and $3$ they made $111,111$. So granny must be $71$!

Neha from Poplar Primary School used a very systematic approach which she describes in detail:

I knew from the start that Mum's age and the kids' ages all multiplied together were $111111$, so what I did was I was about to divide $111111$ by the smallest prime number which is $2$ but I could not because $ 111111$ is an odd number.

So I thought of dividing $111111$ by the second smallest prime number which is $3$ and I could divide $111111$ by $3$ with no remainder. So the answer is $37037$. So I know that one of the kids' age was $3$.

So I thought of the next prime number which is $7$. So I divided $37037$ by $7$ and I could. The answer is $5291$. The next child's age was $7$.

So I thought of the next prime number which is $11$. So I divided $5291$ by $11$ and I could. The answer is $481$. The next child's age was $11$.

So I thought of the next prime number which is $13$. So I divided $481$ by $13$ and I could. The answer is $37$.

So the eldest child's age was $13$ and the mum was $37$!

Then I added all these numbers and got Gran's age which is $71$.

Great work, Neha, although I'm interested that you didn't try dividing by $5$ which is the prime number between $3$ and $7$. I wonder why?

Nikos from Protypa, Thessaliniki, Greece also used a good system, which was slightly different from Neha's. Niko told us:

The number $111111$ comes from multiplication of five numbers. I can see easily that I can divide it by $11$. I do this and get the number $10101$.

This is again a non--even number so I search to divide by numbers like $9$ or $7$ or $5$. I do this and find that $10101=7\times1443$

But again $1443$ is not an even number and I do the same and find $1443=3\times481$

I look what I get up now: $1111111=11\times7\times3\times481$. I again find that $481$ may be divided.

I made tries with $9$ and then $13$ because $5$ or $9$ does not divide $481$ but $13$ can and the result is $37$.

I now have five numbers and the biggest must be the mum=$37$. Then I must add $37+13+11+7+3$ this sum equal $71$. This is age of gran.

Fantastic reasoning, well done Niko. Well done also to Year 4 Maths Class at Hale School and Ella and Helena from North Molton Primary, who used similar methods to Niko and Neha.

Roisin, Charlotte and Jenna from Ribston Hall High School sent us their full reasoning, including explaining what they tried that hadn't worked. Here is their solution:

First of all, we brainstormed some of the different ways to solve the problem:

Trial and Improvement

Algebra

Finding Factors

So, we started with Trial and Improvement by thinking of an age for the mum, e.g. $40$, then some ages for the children, such as $7$, $12$, $14$ and $18$, and multiplied them together until we got $111,111$, because "If you multiply Mum's age with your age and with the ages of your brother and sisters you will get the answer $111,111$."

$40 \times 7 \times 12 \times 14 \times 18 = 846, 720$

This outcome was obviously too high, so we continued with this method and decreased the values as appropriate:

$35 \times 19 \times 12 \times 7 \times 2 = 111, 720$

This is the closest answer we got using this method. We decided this was taking too long and was not the best method to use for this problem.

Next, we tried algebra, but this was also too difficult to use for this problem. We wrote down the expressions, which was useful to know when calculating other sums:

Mum's age ? $m$

Charlie's age ? $c$

Siblings' ages ? $s_{1}, s_{2}, s_{3}$

Gran's age ? $G$

$m + c + s_{1} + s_{2} + s_{3} = G$

$m$ x $c$ x $s_{1}$ x $s_{2}$ x $s_{3}$ = $111111$

Finally, we began using the last method. We worked out what $111,111$ was divisible by and therefore found its factors. The numbers we then had were suitable ages for the mum and the four children:

$1, 3, 7, 11, 13, 21, 33, 37, 39 $

We then chose a combination containing five of the above numbers, one of which had to be a suitable age for the mum and four ages for the children. For example, the mum was $39$; the youngest child was $1$, then $3$, $7$ and $13$.

We used Trial and Improvement to work out what combination was correct by calculating the product of the numbers. When multiplied together $39$, $1$, $3$, $7$ and $13$ produce $10,647$.

The outcome that should be generated if the numbers 'ere right would be $111,111$ because this is the number that is made when the mum?s and children's ages are multiplied together.

After several attempts, we reached the number $111,111$ with this combination:

$37$, $3$, $7$, $11$, $13$

So, when added together the numbers will equal the Gran's age.

$37 + 3 + 7 + 11 + 13 = 71$ years old

Therefore: Gran - $71$ years old

Mum - $37$ years old

Children - $3$ years old, $7$ years old (Charlie because he is the third child), $11$ years old, $13$ years old.

The Finding the Factors method was the easiest, quickest and simplest way to solve this problem, because after a few minutes of starting to use that method, we had the answer. It was a logical way of answering the problem, rather than using Trial and Improvement, however, it was also straightforward, whereas using algebra proved to be too difficult.

What a wonderful description of your approach.

Laina from St John the Baptist's School in Shropshire found a different solution. Here it is:

37x 11x21x13x1= 111,111

37+11+21+13+1=83

So gran could be 83 years old as well. She is old!

Can you think of any problems with this solution? Thank you for sharing your ideas with us and thank you to everyone who sent in a solution to this problem.

Gran, How Old Are You?

How old was his Gran?

Why do this problem?

This problem involves unknowns, and encourages algebraic thinking, but does not rely on using letters to represent the unknowns. Instead, it encourages multiple approaches, so learners might use trial and improvement, for example. It is an ideal opportunity to reinforce the idea that there isn't just one way to solve a problem, although some methods might end up being more efficient than others (for most people). This challenge also consolidates understanding of multiplication and division being inverses of each other, and it might offer a chance to discuss divisibility rules.

Possible approach

Introduce the problem to the class as it stands, perhaps orally, or by projecting the page onto the whiteboard. Give everyone time to think about how they might approach it in pairs or small groups, being very careful not to say anything else at all. This may feel very uncomfortable for some children (and for you!) but try not to help them at this stage.

After a suitable length of time, bring everyone together and invite comments on how the problem might be tackled. Once again, try not to advocate one method but once several ways have been discussed, explain that you will give more time for working on the solution now. Each group can choose how they go about the problem so they could use a method that someone else has suggested.

After more time, you could bring the whole class together again to talk about progress so far. Then allow a period of time for further work on the solution and perhaps also for each pair/group to summarise their method/s on a poster, including reflection on how well their method/s worked.

The last part of the lesson could be used for groups to look at each other's posters and compare solutions. This could lead into a discussion about the advantages and disadvantages of each method.

Key questions

What information do you know?

How could you use this to try and find a solution?

What could you try first?

How old do you think Charlie's mother could be?

How old do you think Charlie could be?

Possible extension

Learners could find numbers with interesting factors and make up their own similar problem. Does their own problem have a unique solution? How do they know?

Possible support

Some children may find a calculator useful.

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