Mystery Matrix


I have to say that I am very impressed by the number and quality of the solutions you have sent in to this Mystery Matrix problem. I am afraid I can't mention everyone individually but special mention must go to pupils at Wharncliffe Side Primary School, Stourport Primary School, How Wood Primary and Mount Pleasant Juniors, all of whom sent in well-explained solutions.

Rukmini from Roberts Elementary School wrote:

This is my solution to the Mystery Matrix:

I first realized that if you put your finger on a square and track its row and column, the number in the square would be the product of the two numbers - the one at the head of the row and the one at the head of the column.

I knew that each column was part of a multiplication table. I realized which numbers were needed to multiply to get that product;

e.g. $32$ was a given number and I knew that $4 \times 8 = 32$, so I put $4$ and $8$ as the heads of the row and the column. $8$ needed to go at the head of the column and $4$ needed to go at the head of the row. $40$ was a given number and the head of its column was $10$. That helped me decide where to put the $4$ and $8$. If I switched $4$ and $8$ then I wouldn't be able to track $4 \times 10$ to $40$. I knew where to put the $4$ and $10$ because if I switched them I wouldn't be able to track the $4$ and the $8$ to $32$.

$49$ was the number I tackled first. I knew that $7 \times 7$ was $49$. So it didn't matter which $7$ I put at the head of the row or the head of the column.

Then I did $32$ as I already explained.

Then $7 \times 8 = 56$, so I put $56$ in the blank box tracked from heads $7$ and $8$.

Using the rest of the given numbers I figured out what the rest of the heads were. I used the heads to help me figure out what the rest of the numbers were.

Mrs Beacham's class from Round Top Elementary School, Blythewood, USA also sent a very clear account of the way they solved the problem with a picture of the finished matrix:


First, we came up with the factors for each number in the matrix.
We looked at the hint and figured out that $7$ had to be the repeated digit, because there is only one way to get $49$.
Then we compared the numbers in each row and column where there were two numbers. For the numbers, $22$ and $24$, $15$ and $27$, $49$ and $42$ we realized they only had one factor in common. When we compared $22$ and $24$ we determined that they were both multiples of two. So $2$ had to go in this column, which meant that $11$ had to be the missing factor for $22$, and $12$ had to be the missing factor for $24$. $15$ and $27$ share the factor $3$, so we placed a $3$ in that row, and above the $15$ we placed a $5$, above the $27$ we placed a $9$.
For $49$ and $42$, we already knew they shared the factor $7$, so we thought what times $7$ will give us $42$, and that was $6$.
The only digits left to use in the matrix were $4$, $8$, and $10$. When we started, $32$ and $40$ were the hardest to figure out because these numbers shared more than one factor. If we put the factor $8$ in the row, that would have meant using the factor $5$ again, so we realized that the factor $4$ had to be placed in that row. $8$ had to go above the $32$, and we placed $10$ above the $40$.
Finally, we multiplied the factors in each row and column to come up with the missing products.

Anhar from Holy Cross Primary School, Oldham, also sent a very clear explanation of the steps he followed to get to a solution which are in a slightly different order compared with those of Mrs Beacham's class:


The first thing I did was work out that $22$ and $24$ are both in the $2$ times table. I put $2$ at the top of the third column.
The second thing I did was notice that $49$ and $42$ are both in the $7$ times table. So I put $7$ at the top of the fifth column.
The next thing I noticed was that $15$ and $27$ were both in the $3$ times table. I put $3$ in the fourth row down.
Now I have worked out that $7$ is the fifth column, I know it must be $7 \times 6$ that equals $42$. So $6$ goes into the bottom row.
Using the same method we knew $7 \times 7$ equals $49$, so we put $7$ in the second row down. This also meant that $7$ had to be the number we used twice.
Now that we know the third column is $2$, we can work out that $2 \times 11$ equals $22$ and $2 \times 12$ equals $24$. So $11$ goes into the third row and $12$ goes into the fifth row.
Knowing $3$ is the fourth row down, we can work out that $3 \times 9$ equals $27$. So I put $9$ in the last column.
We also worked out that $3 \times 5$ equals $15$, and placed $5$ in the second column.
We then worked out that $12 \times 2 = 24$, so I placed $2$ in the third column.
I knew that $8 \times 4 = 32$. However, I did not know which to put in the column and which in to the row. I then tried putting $8$ in the column and $4$ in the row. But that would have meant putting a $5$ in the remaining column, meaning I would have used $5$ twice as well as $7$ twice. I switched the numbers around, $4$ to the column and $8$ to the row.
I placed $10$ in the remaining column and completed the number matrix.

Fantastic! Thank you, Anhar. Finally, Juliet and Lauren from Whipton Barton Junior School sent in a solution in which they made it clear that they were using a 'trial and improvement' approach, which can be a very powerful way of going about solving a problem like this one. Here is what they wrote:

We solved this problem by working out what could go down the sides.
First of all we worked with $32$. We did this by thinking of factors of $32$ and we found $8$ and $4$. We then thought about other factors of the numbers on the grid. It was pot luck whether we got the numbers the right way round along the sides at first, but if we noticed a problem we could change it.
We noticed that $49$ was a square number and we knew that the two factors timesed together was $7$ times $7$. This meant $7$ was the only number that appeared twice.
Also, we could fill the middle of the grid in using our times tables facts.

Once again, thank you to everyone who sent in a solution which explained your method.