Add to 200
1 out of 3
By selecting digits for an addition grid, what targets can you make?
Problem
Choose any four digits (from 0, 1, 2, ... , 9) and place them in the cells below (you can repeat digits).
Read the two 2-digit numbers across and add them together.
Read the two 2-digit numbers down and add them too.
Then add these two totals together.
For example:
Here we got a total of 125.
Try a few examples of your own. What totals can you make?
Is there a quick way to tell if the total is going to be even or odd?
What happens to the total when you add 1 to the top left cell? ... to the top right cell? ... to each of the bottom cells?
Challenge
Can you make a total of 200?
Can you find all the ways of doing this?
How could you convince someone that you've found them all?
Possible extension
If we place 0 in every cell, the total would be 0.
If we place a 5 in the top left cell and 0s in every other cell, the total would be 100.
Is it possible to make all the totals between 0 and 100?
With thanks to Don Steward, whose ideas formed the basis of this problem.
Getting Started
If you start with
| $A$ | $B$ |
| $C$ | $D$ |
How will the total change if you increase $A$ by $1$?
How will the total change if you increase $B$ by $1$?
How will the total change if you increase $C$ by $1$?
How will the total change if you increase $D$ by $1$?
How can you ensure that the total is going to be even?
Student Solutions
Is there a quick way to tell if the total is going to be even or odd?
Ritun from Heckmondwike Grammar School in the UK, Shaunak from Ganit Manthan in India and Thomas from Dulwich College in the UK described the same method as each other. Ritun wrote:
You can take the top-right number and the bottom-left number, bottom-right number twice on the grid and add them up, if they are odd when added the final number will be odd, if they add up to even then the final number will even.
For example :
$1+6+6+2=15,$ $15$ is odd so the final number will be odd and it is, $125.$
Grace and Emily from Long Field Spencer Academy in the UK and the Harriets from Colyton Grammar School in the UK had a shorter method. This is Emily's work:
A quick way to know if our total will be odd or even is by adding the top left and the bottom right numbers together. Example:
Bethany from the UK thought about what happens when each number is odd or even. Although Bethany doesn't write down the rule, you can see it in the working. Here is Bethany's work:
Can you make a total of 200?
Mabel and Florence from Long Field Spencer Academy both found numbers that gave totals close to 200. Florence wrote:
The closest I found was $\begin{array}{cc} 9 & 1 \\ 1 & 0 \end{array}$ the outcome of this is 202, I couldn't get any closer.
Mabel wrote:
The closest answer to 200 that I could find was $\begin{array}{cc} 9 & 0 \\ 1 & 4 \end{array}$ and $\begin{array}{cc} 9 & 0 \\ 1 & 5 \end{array}$ these two solutions reach an answer of 199 and 201.
Lots of people did find combinations that added up to 200. Year 10 from Hautlieu School in Jersey, Daniel from Outwood Academy Hasland Hall in England, Daniel and Sophia from Kingsway School in New Zealand, Rubi from Nottingham Girls High School in England, Will and Vic from Colyton Grammar School, Kevin, Gabirel, and Max from St John's College School in the UK, Fatema from Archbishop Blanch School in England, Vardhini from Comberton Village College in the UK, Ted from Whitefriars Academy in the UK and Javier from Twyford in the UK all found the same array. Fatema wrote:
To make a total of 200 you must put in these numbers:
4 5
5 5
45 + 55 = 100
45 + 55 = 100
100 + 100 = 200
Ella Mae from Long Field Spencer Academy, Adam and Daniel from Outwood Academy Hasland Hall, Vardhini, Ted and Javier found another solution. This is Ella Mae's work:
Phoebe from Long Field Spencer Academy, Harry from St Johns College School, Rubi, Vardhini and Shaunak found another solution. Phoebe wrote:
If you have 72 along the top and 28 along the bottom then you can solve this problem.
Horizontally:72+28=100
Vertically:72+28=100,
and 100+100=200
Mei Mei from St Johns College School, Rubi, Vardhini, Ted and Javier found another solution. This is Rubi's work:
Year 10 from Hautlieu School, Daniel from Outwood Academy Hasland Hall, Vardhini and Ted found another solution. Vardhini wrote:
27
73
Daniel from Outwood Academy Hasland Hall, Nathan and Jules from Colyton Grammar School, Ted found another solution. Daniel wrote:
18,82
Daniel from Outwood Academy Hasland Hall, Year 10 from Hautlieu School, Ted found another solution. This is Year 10 from Hautlieu School's work:
Rubi, Vardhini, Ted found two more solutions. This is Ted's work:
Mia from Long Field Spencer Academy and Ted found another solution. Mia wrote:
4 / 6
-----
4 / 5
45 + 46 + 44 + 65 = 200
Below are the solutions that were only found by one person or group. This is from Taylor from Long Field Spencer Academy:
Summer from Long Field Spencer Academy:
I guessed random numbers to put in the boxes and then added the numbers together e.g. 29 + 53 = 82 then I did 93 + 25 = 118 finally I added these two totals together and got 200
Tommie from Long Field Spencer Academy:
Isla from Long Field Spencer Academy:
Nathan and Jules:
Ben and Emily from Portobello High School in Scotland:
Shaunak:
Some people described their methods. Daniel from Outwood Academy Hasland Hall, Arthur and Lorcan from St John's College School, Fifi from Djibouti and Javier used the same number twice. Arthur wrote:
Your bottom left square and your top right square [are] opposite numbers. [If they're] the same and then figure out the other two so both numbers on each side [add up] to equal one hundred so then they will eventually all add up to 200. Like the the bottom left and top right are 4 and then the top left is five and the bottom right is 6 so they both equal one hundred because 54+46 equals one hundred and then double it.
Javier and Fifi expressed this idea using letters to represent the numbers. Javier wrote:
b = c
Freya and Emily from Long Field Spencer Academy, Sofia, Sophia and Issy and Louis from Colyton Grammar School, Bethany, Fifi, Ted, and Javier used the row and column totals. This is Bethany's work:
Freya and Emily used the row and column totals idea to generate 9 solutions. They are all shown above, but click to see them all together in Freya's spreadsheet.
Javier and Fifi additionally noticed that:
d = a + 1
The methods above all find solutions where both pairs of numbers add up to 100. We know from the examples above that there are also solutions where this is not the case. The methods below also find these solutions.
Isabelle from Long Field Spencer Academy thought about the place value that each number could have:
Yuan from Colyton Grammar School also used this idea, and described how to find another solution, if you've already got one:
Increasing top-left and bottom-right by 1 will increase the final result by 22 because top-left is the tens digit for both its sums, and bottom-right is the units digit for both its sums. Therefore increases both by one will increase it by (10 $\times$ 2) + (1 $\times$ 2), which is 22.
Decreasing top-right and bottom-left by 1 will decrease the final result by 22. This is because both top-right and bottom-left are the units of one of its sum and the tens of the other. Decreasing the top-right and bottom-left will each decrease the final result by 11 and so both will decrease it by 22.
In this logic, by increasing top-left and bottom-right, and decreasing top-right and bottom-left will increase the result by 22 and decrease it by 22, thus cancelling each other out and leaving the answer of 200 the same, despite changing the numbers.
Year 10 from Hautlieu School, Thomas from Dulwich College and Ahana, Sehar, Saanvi, Dhanvin, Aariz, Ananthjith, Vivaan, Sai, Pranathi, Paavani, Utkarsh, Nikhil, Dhruv and Shubhangee from Ganit Kreeda in India expressed the calculations using algebra. The Ganit Kreeda students wrote:
So, Total $= 20a + 11(b + c) + 2d$
This enabled them to fully explain the rule to determine whether the result is odd or even that Grace, Emily and the Harriets described:
The digit at b & c is used as ten’s place digit and unit’s place digit alternately.
The digit at d is always used as units place digit.
So, b + d + c + d = b + c + 2d will decide the digit at unit’s place from the total.
Now 2d is always even which means,
the total is even if b + c is even (b + c can be even if both b & c are even or if both b & c are odd.)
and
the total is odd if b + c is odd (if one is even and other is odd).
The algebraic representation allowed Thomas to describe a strategy for finding solutions:
Using this you can go down in twenties from 200 until you find a number you can make with elevens and even numbers under twenty.
Year 10 from Hautlieu School wanted to find all the possible solutions:
$20x+11A+11B+2Y=200$
After this it was then obvious that $11(A+B)$ meant that $A + B$ needed to be even as $20x, 2Y$ and $200$ are even.
We then all worked on fixing the top left number as $0, 1, …, 8$ and looking for even multiples of $11$ that the $Y$ would be a number from $0$ to $8$ which makes $11(A+B) + 2Y = 200 – 20x$ work.
For example for $5$ in the top left corner, we noticed that $20\times5+11(A+B)+2Y=200.$
We needed $11(A+B)+2Y=100.$ The even multiples of $11$ close to $100$ are $88$ and $66,$ but $66$ is too far away ($Y$ would need to be $17$ which it cannot be).
$11\times8+2Y=100,$ so $Y=6.$ This gave us:
| $5$ | $B$ |
| $A$ | $6$ |
Where $A+B=8.$
| $A$ | $B$ |
| $0$ | $8$ |
| $8$ | $0$ |
| $1$ | $7$ |
| $7$ | $1$ |
| $2$ | $6$ |
| $6$ | $2$ |
| $3$ | $5$ |
| $5$ | $3$ |
| $4$ | $4$ |
And this gave all the possibilities for 5 in the top left hand corner.
This [is our final collaborative work]:
Which isn’t complete but our lesson ended! And we knew what the idea was that would fill out the rest.
The students from Ganit Kreeda used a very similar approach:
As we want total to be 200, we need ‘0’ at unit’s place, which means $b+c$ is even and $b+c+2d$ must end in 0/10/20/30.
$b+c$ can take all even values from 0 to 18
$20a + 11(b + c) + 2d = 200$
All the kids from the Ganit Kreeda worked collaboratively by putting all the values for $b+c$ to find out all possible solutions using above strategy and algebra:
Dan from Long Field Spencer Academy wrote a program which finds all possible solutions:
I created some code in python that utilizes algebra by finding out the algebraic solution to the problem. This code took about 10 minutes, but I tested some of [the solutions] to see if they worked and they did.
Click to see all of the solutions outputted by Dan's code.
The leftmost number is the top left number in the box, then it's top right, then bottom left, and finally bottom right.
The rest of the solutions are, 3 5 7 4
3 6 6 4
3 7 5 4
3 8 4 4
3 9 3 4
4 1 9 5
4 2 8 5
4 3 7 5
4 4 6 5
4 5 5 5
4 6 4 5
4 7 3 5
4 8 2 5
4 9 1 5
5 0 8 6
5 1 7 6
5 2 6 6
5 3 5 6
5 4 4 6
5 5 3 6
5 6 2 6
5 7 1 6
5 8 0 6
6 0 6 7
6 1 5 7
6 2 4 7
6 3 3 7
6 4 2 7
6 5 1 7
6 6 0 7
7 0 4 8
7 1 3 8
7 2 2 8
7 3 1 8
7 4 0 8
8 0 2 9
8 1 1 9
8 2 0 9
Which numbers between 0 and 396 is it possible to make?
Sofia, Sophia and Issy focused on several specific numbers:
100
To get 100, the digits in the left hand column have to add up to 4 and the digits in the right hand column have to add up to 10 and the diagonal column from bottom left to top right have to be the same.
150
To get 150, the digits in the left hand column have to add up to 5 and the digits in the right hand column have to add up to 7 and the diagonal column from bottom left to top right have to be the same.
300
To get 300, the digits in the left hand column have to add up to 14 and the digits in the right hand column have to add up to 10 and the diagonal column from bottom left to top right have to be the same.
Bethany worked out which numbers can be made by starting with four copies of the same digit:
Teachers' Resources
Using NRICH Tasks Richly describes ways in which teachers and learners can work with NRICH tasks in the classroom.
Why do this problem?
Depending on students' experience of algebra, this problem also offers an opportunity for introducing how algebra can be used to simplify the problem and shed light on how to find the total number of solutions. This may be less familiar to students than other ways in which they have used algebra.
Possible approach
Start with the grid drawn on the board and ask students to copy it into their books.
Ask them to fill their grid with four digits of their choice. Ask them to read the two 2-digit numbers across and the two 2-digit numbers down, and then add them altogether.
"What totals did you get?"
List some of their totals.
"What questions might a mathematician ask now?"
Some fruitful questions are:
When are the totals even and when are they odd?
Which totals between 20 and 30 can you make?
What happens when you increase one of the cells by 1?
Can some totals be made in more than one way?
Allow students to work together on a question of their choice.
After some time, bring the group together and discuss any insights they have gained. This should prepare them for the main challenge:
Is it possible to make a total of 200? Can they find all possible ways?
Members of the NRICH team have worked with students on making all the possible combinations that add to 200. As the students found each combination, we displayed it on the wall in a very particular arrangement, but without explaining what the arrangement was. As this 'jigsaw puzzle' was building up, students were able to identify missing solutions, and eventually started placing new solutions in the correct place themselves. The arrangement facilitated discussions about the patterns down a column and across a row, and why these patterns occur. Here is an image of the final arrangement of all the combinations (including an interesting one on the far right!):
After students have found all the combinations that give 200, you could bring the class together to consider how they could gain further insights by using an algebraic approach.
If we start with
| $A$ | $B$ |
| $C$ | $D$ |
the total can be written as $20A+11B+11C+2D=20A+11(B+C)+2D$.
So we need $20A+11(B+C)+2D=200$.
This looks a bit daunting as it has four unknowns, but we can make it simpler by fixing a value for $A$:
For example, if $A=3$ we need $11(B+C)+2D=140$.
We know that $2D<20$, and even, so we need $120<11(B+C)<140$.
$132$ is the only even multiple of $11$ between $120$ and $140$ and $132=11\times 12$, so $B+C=12$ (see the 4th column in the 'jigsaw puzzle' above).
Students could work systematically to predict what happens for each possible value of $A$ and check this with the completed 'jigsaw puzzle'.
Add to 200 could be used as an activity for students learning to program. They could write a program to find all possible solutions.
For example, this program was written by Nevil Hopley, Head of Mathematics at George Watson's College, for the TI-Nspire software.
If you have the appropriate software, you can download his .tns file here: Add to 200.tns
Key questions
When are the totals even and when are they odd?
What happens when you increase one of the cells by 1?
Can some totals be made in more than one way?
Possible support
Having digit cards available for students to physically manipulate or mini whiteboards will help those who are reluctant to commit ideas to paper.
Possible extension