Dicey Operations

Adding two two-digit numbers

The children at Copthorne Prep School in the UK thought hard about strategies for this game. Alexander tried to find a pair of numbers that added to 10 to put in the tens column:

I worked out that if two numbers were ten if added together if you add them in the tens column and then the other two in the units columns they would be close to 100.

The children from St Margaret's School for Girls in Scotland had some similar ideas. Abbey explained:

I use number bonds. I also put the biggest number first if it doesn't go too far over 100. For example if I had 7, 4, 9, 2 I would put 29 together and 74 together because if I put 9 and 7 first I would go over 100 by a lot.

Tiamike looked for digits that added to 9 rather than 10 for the tens column:

When I got my four numbers for example 8,4,1 and 5 I thought what makes ninety. I knew that 8 + 4 would go over and 1 + 5 was a little too low so then I added 8 + 1 giving me 9 then I had 4 and 5 left over and I knew 4 and 5 also made 9 so I added the answers together so I was left with 99. That was the closest thing I could get to 100. I also used my number bond knowledge to help me and a bit of trial and error as well.

The children from St. Martin's Catholic Primary in the UK had some similar ideas. Myla and Otto explained:

We tried to make the ones add up to 10 and the tens add up to 90.

Kush from Doha College in Qatar also looked for numbers that added to 90, and explained why it was important to think about the tens column before the ones column:

Try to do the tens place first, as this will be easier to get close to 100, then do the ones place last. For example, if you have the digits 4, 2, 7 and 5, place the 5 and 4 in the tens place to get 90, then place the 2 and 7 in the ones place to get 99.

Noah from St Augustine's Catholic Primary School in the UK explained why sometimes, it's even a good idea to look for tens digits that add up to 80 rather than 90:

If you have 5,7,3, and 4 as your numbers you choose 5 and 4 as your tens digits because 50+40 is closer to 100 than 50+30 or 50+70. Why don't we do 70+30? Because of the ones digits. The result we would get from using 50+40 would give us 50+40+7+3=100 but using 70+30 would give us 70+30+5+4=109. If you have 2 digits that add to more than 5 but less than 16, you should try to get as close to 90 as possible with your remaining digits. If they add to 16 or above, aim for 80 with your 2 other digits. If they add to 5 or less, aim for 100 with your 2 remaining digits. 5+4=9 which is between 5 and 16. Therefore you should aim with 90 with your remaining digits, not 100.

Adding three three-digit numbers

The children from Banstead Prep School in England worked on the version of this problem where three-digit numbers were added together. Avyu tried to make the 100s column add up to 900, similarly to Tiamike's strategy above:

I decided where to put my digits by starting in the hundreds going down to the ones.

My strategy was making sure the 100s column adds up to 9 and the other columns 10 or close to it because it will make the number close to 1000.

Jasper noticed the same thing as Noah, above - sometimes if the numbers are large, making an 8 in the largest column is the best strategy:

I decide where to put my digits based off roughly making a ‘9’ in each column so with exceptions because it should total to near 1000. If I have high numbers for example, I will try to make 800 in the 100s column because I have a larger gap to fill in with my higher numbers.

My strategy is to always start with the hundreds and work round the type of numbers I have (e.g. low, high).

The students from William Konkin Elementary School in Canada had some similar ideas. Aubree and Sienna explained their strategy for adding to 9 in the hundreds column:

Oscar and Nelson built on this strategy, explaining why sometimes adding to 8 in the hundreds column can get the answer closer to 1000:

Hank from Stowe School in the UK realised that sometimes, choosing hundreds digits that only add up to 7 can work well:

1. This time need to try out multiple possible combinations for the hundreds digits of the three numbers.
2. Start from the combinations of three numbers that added up to 9 as the hundreds digits for the three numbers. Then this question is reduced to find three double-digit numbers with summation as close to 100 as possible.
3. You can try out the combinations of three numbers that added up to 10 as well. But generally it will not be better than 9.
4. Then try out the combinations of three numbers that added up to 8. This question is reduced to find three double-digit numbers with summation as close to 200 as possible.
5. Generally, a combination of three numbers that added up to 7 is unlikely to give better results when three are possible combinations for 8 and 9. But it is not impossible. Only try combinations of three numbers that added up to 6 or less when summation as 7, 8, or 9 is not possible.
6. Only try out combinations of three numbers that added up to 11 or more when there is no possible combinations of three numbers that added up to 10 or less.
7. Can fine-tune the result according to the relationship between the current summation and the target.
8. Find the scenario that has the closest summation to 1000.

Subtracting

Hank developed a method for each of the three subtraction versions of the game:

Sub 1

1. Try the pair of digits that are consecutive as the tens digit of the two numbers first.
2. Then try the pair of digits that are the same as the tens digit of the two numbers if a pair of same number is available.
3. Then try the pair of digits with difference of 2. When filling in the ones digits, fill the larger number into the ones digit of the number below.
4. Only try other scenarios when none of the scenarios described above are possible.
5. Find the scenario that has the closest result to 10.

Sub 2

1. Try all the combinations of the two numbers with difference as 1 to be the hundreds digits of the two numbers. Then this question is reduced to finding two double-digit numbers that are the closest to each other with the rest of the numbers.
2. Then try all the combinations of the two numbers that are equal (if possible). Then this question is reduced to finding two double-digit numbers that are further apart as possible with the rest of the numbers.
3. Only try other possible combinations when you can’t get the result from the previous scenarios to be within 100 of 100.
4. Can fine-tune the result based on the result’s relationship with 100.
5. Find the scenario that has the closest result to 100.

Sub 3

1. First look at all the available numbers and determine whether there are repeated numbers.
2. If there are, first try to take one pair of equal numbers as the hundred digits of the two numbers. Then try a pair of consecutive numbers as the thousands digits. Finally try to find two double-digit numbers that are closest to each other with the numbers left. Next try a pair of equal numbers (if possible) as the thousands digits. Then try to find two double-digit numbers that are as further apart to each other with the numbers left. If you cannot get the result within 100 of 1000, try other combinations for the thousands digits and hundreds digits. In the end find the scenario with the closest result to 1000.
3. If there is not yet a solution. Just follow the same procedure as Sub 2 with an extra set of operations for the thousands digits.

Dhruv from The Glasgow Academy in Scotland tried to make the largest total for the four-digit subtraction version:

If we are trying to get the highest total we need to have the biggest number on the left side and the smallest number on the right side.

Example

The digits I got are 2, 3, 8, 1, 6, 9, 1, 5. To make the largest number on the left you need to put the digits in descending order from left to right.

To make the smallest number on the right you need to put the digits in ascending order from left to right.

So the way we need to put the digits for the example are 9865 - 1123 which equals 8742 which is the highest total for these digits.

The children from Twyford School, Winchester, thought carefully about every version of this problem. Take a look at Twyford School's full solution to see their ideas.

Printable version

Some children spent a lot of time working on the version of the game which involves rolling dice and deciding where to place the digits one at a time.

Roman, Elliot and Jake from Central School Te Kura Waenga o Ngāmotu in New Zealand had a strategy for where to put numbers of different sizes in the addition grid:

When we roll a number six or below we like to put it in the first box in the hundreds and anything 3 or below we put in the next two boxes in the hundreds place. Anything 1 through 8 we put in the tens and any number in the ones.

Elli and Sofia from Shatin Junior School ESF in Hong Kong thought hard about different strategies they could use when playing this version of the game. Take a look at Elli's solution and Sofia's solution to see their ideas about this.

General insights

Christopher from King George V School in Hong Kong noticed something very interesting about the different possible totals it was possible to make with a single set of numbers:

First of all, suppose we have a 2×2 array to fill in, we have the digits a,b,c,d, and the target number is t. If we have ab and cd as our 2 2-digit numbers, it probably won't be the closest you can get to t. Notice that when we swap ab with ba, we increase our sum by 9(b-a) since ba - ab= 10a+b-(10b+a)=9(b-a). We also find that if we swap a and c so that the 2 2-digit numbers are ad and cb, the sum is still the sum. Since every possible combination of 2 2-digit numbers can be obtained by a sequence of swapping a ones digit with a tens digit, every possible sum differs to every other sum by a multiple of 9.

If you are confused already, here's an example: suppose the 4 digits are 3,5,2,8 and the
target is 100. We start by just using a random pair of 2-digit numbers: 23 and 58. They sum
to be 81 so we know that the closest we can ever hope to get is 99 because it is the closest
number to 100 that is a multiple of 9 away from 81. We do some guesswork and swapping
and find out that 99+9=108 is the closest we can get by just using 23 and 85 which sum to be
108.

Take a look at Christopher's full solution to see more ideas about this. This is a very powerful insight, because it makes it much easier for us to work out whether or not our solution is as close to 100 as it is possible to get.

We received a lot of solutions to this problem, and unfortunately we don't have space to share them all here. The following children sent in similar solutions: Marvin from St. Mary's International School in Japan; Ayah from Doha college in Qatar; Advait from Kings' School Al Barsha in the UAE; V from The Camford International School in India; Shivaprasad and Valan from The GYM Foundation in India; David from the American International School of Budapest in Hungary; Vihaan from Singapore American School in Singapore; Amelia from Doha College in Qatar; Hector from Renaissance College Hong Kong; Syan from England; Arvin from Mountain View Academy in Canada; and Shreehari from Ganit Kreeda in India. Thank you all for sharing your ideas with us.