Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position
yourself so that you are 'it' if there are two players? Three
Can you order the digits from 1-6 to make a number which is
divisible by 6 so when the last digit is removed it becomes a
5-figure number divisible by 5, and so on?
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:
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
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.