Dicey Operations

The children at Copthorne Prep School in the UK thought hard about strategies for this game. Alexander said:

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.

V from The Camford International School in India said:

First, you need to pick two numbers from the given four that add up closest to ten.

The two numbers should then be placed in the tens places.

The other two numbers will be placed in the ones place.

These should be placed in such a way that the numbers' sum is as close to 100 as possible.

Hector from Renaissance College Hong Kong said:
Addition solutions:

2 digit by 2 digit solution:

To find the optimal answer for these problems, you first have to analyse the largest place currently available. For example, if you are doing the 'closest to 100' challenge, the priority lands on the tens place first. Depending on the numbers you get, the formula will adapt. Here are the 2 most effective solutions for this problem:

1) Since you are adding 2 numbers together to reach closer to 100, the 2 numbers in the tens place could not drastically be stretched far apart. The 2 numbers has to reach 10 as close as possible. For example, if you have the numbers 4, 6, 7, and 8. So far, only the numbers (4 and 6) make sense, because while added up gives a total of 10. Any other combination is not as close, which rather falls under or goes over.

You can compare it like this:

4+6 = 10

7 + 8 = 15

6 + 8 = 14

Overall, the first option is closest to 10. The rest of the numbers could be placed in the remaining areas. Finally,  the rest of the numbers has to be checked. If the outcome for the ones ends up higher than 5, its better for the tens number to add up to a 9. Since in this instance, a 96 is closer to 100 than a 106.

2) This is the second condition for this problem. Sometimes the closest number to 10 is not the best solution. For example, the numbers 8, 9, 1, 2. Usually,  the first formula would work really well. But in this case, it is not the most optimal. If you adapt to the earlier method, 8+2=10, and 9+1=10. Both of these work right? But actually,  this trick leaves another flaw. Lets say the 8 and the 2 go into the tens place, giving remainder to the 1 and 9.

 81

+29

------

110

Looking at this, not only does the 8 and 2 make 100, the 9 and 1 makes a 10. This gives 110.

If you ever come across a problem where the ones place adds up to become a double digit number, you have to adapt. Now the new rule is how close you can get the number to 9. This leaves enough space for the ones to carry into the tens, allowing for it to be closer to 100.

8+1=9, which leaves 2 and 9 behind. After adding them up, it will become 101. because 2+9=11, the extra ten carries into the 90(from 80+10), which creates a full 100.  Another special occasion is when the ones don't add up to a full 10, but is within range of it. For example, it could add up to 8 or 9. This is most likely the correct answer for the problem to use a 90+ a number close to 10.

3) The third and final condition is the ones place numbers. If you end up receiving high numbers like 9 and 8, added together would give you a 17, where rounded up is around a 20. This is a special case, where you try to drop it to a closer 8. For example, the numbers 1, 7, 8, 9. Right now there are 3 large numbers, and when added together would guarantee a rounding towards 20.  In this instance, don't add the closest number to 10(like 9 and 1), but because 3 of our numbers if paired in any way gives a rounding for 20, go for an 8 combination, like 7 and 1. 

The children from St Margaret's School for Girls in Scotland thought hard about this. 

Meagan said:

I do trial and error and I mostly put the bigger numbers in the units column.

Evangeline said:

I did trial and error. I saw the biggest number and then added smaller number and tried to get the closest number to a hundred.

Abbey said:

I use number bonds. I also put the biggest number first if it doesn't go to far over 100. For example if I had 7492 I would put 29 together and 74 together because if I put 9 and 7 first I would go over 100 by a lot.

Milena said:

First I looked for the smallest and biggest number and added them then put them in the tens and put the other two in the ones.

Ella said:

I looked for number bonds and added them together to get to a hundred 

Izzy said:

For dicey operations I did the process of elimination first I tried every possible answer, wrote them down and added them all. I submitted the closest to 100.

Tiamike said:

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.

Noah from St Augustine's Catholic Primary School in the UK said:

Let's say you are chosen 4 random digits between 1 and 9. To get the sum close to 100, you use number bonds. 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.

Kush from Doha College in Qatar said:

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.

Amelia from Doha College in Qatar said:

The numbers given were 5,0,5,1 and I first tackled this by finding a two digit number as this is what the question is looking for. I realized that 05 and 15 = 20 and would be too small compared to 100 and too far from the target number I wanted to reach - 100. I then also put together 55 and 10 = 65, which would be somewhat closer than adding 55 and 01 =  56 together. I knew that there could be one more method to get closer to 100, so I matched the numbers 50 and 51 together and when added they sum to 101. So in conclusion, the number 101 which is 51 + 50 is the answer where two numbers that are two digits are closest to 100.

I first looked at the numbers I was given - 4,0,6,8 and I started by eliminating options that were too far from the number I wanted to reach - 100. I started by adding 68 and 04, which gave me 72, and then I also added 86 and 04 which gave me 90. I then also tried adding the numbers 68 and 40 and I got 108, which in conclusion was the number closest to 100, with 8 as the remaining number.

David from the American International School of Budapest in Hungary said:

 69
+34
-----
103

Because putting 6 and 4 in the tens place would already put us over 100, there is still 13 left. Therefore, we have to remove from the tens place to have another value go into the ones digit place. Logically, it would seem that removing the 4 would be good since the 6 gives us a fair value of 9 to stay at, thus, that already implies we need a 9 in the tens place since our other numbers will add up themselves and go over 100 but by a smaller value.

The children from Banstead Prep School in England had lots of ideas.

Avyu said:

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.

Zain said:

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My strategy is to try and begin with 9 as a starter, then eight then 7 etc (in the hundreds), because then I can use that to figure the tens and ones digits

Jasper said:

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).

It is hard to get to the target if you do not have a large variety of numbers.

The closest I got to 1000 is 2 away.

Daniel said:

It was more difficult when the numbers were bigger. My solution is to do the hundreds first and get rather 9 or 8 in total. Then I would move on to the tens an then the ones.

Naomi said:

How close can you get to the target?

I could get 6 away from 1000 (1006) which in this question was the closest you could get.

How are you deciding here to put the digits?

I decide where to put the digits by starting with the hundreds and making sure that in total it adds up to 9. Then in the tens column I make sure that it adds up to 10. Sometimes this does not work as the numbers in the tens column do not always add up to 10 so I have to either find another way to make the 100s nine or make the 10s as close to ten as I can.

Will my strategy always work?

My strategy wont always work as sometimes you have too many big numbers and you have to get as close to nine and as close to ten and that could be five away.

When did my strategy work best?

My strategy works best when I have lots of small numbers and zeros to make 9 and 10.

When did my strategy not work well?

My strategy didn’t work so well when I had barely any zeros to work with because I had to go over nine and ten.

Cooper said:

The closest I have got to the target so far is 1004 with 5,3,1,5,4,0,4,9 and 1. I decide where to put the digits by starting at the hundreds then the tens then the ones. My strategy is to think about what numbers I have left and see if it works for the others before I put the digits in. This strategy normally works when you have lots of different numbers ranging from low to high including zero’s. It is hard if you have lots of big numbers. It is perfect if the ones add up to 10 and the tens and hundreds add up to 9. Lots of zeros will help you because when you add three numbers they all have to be 3 or less than 3 to not carry a remainder which is what you want for the hundreds and tens. You will need 0,3,6,4,4,1,0,7 and 3 to make exactly 1000 (or anything that adds up to 28)

Elliot from England said:

I decide where to put my digits by trying to have a total of 800 or 900 in the first column if it is possible to do so. If it is not possible then you must have the smallest numbers in the hundred's column and the biggest numbers in the one's column. Here is an example of my strategy if it is not possible to have 800 or 900 in the hundred's column.

If it is possible to make 900 or 800 in the hundred's column then you can have around 90 in the ten's column and around 10 in the one's column. Try and work around this trick if it is not possible because you might not get good enough numbers to use this trick. The closest I got was 1002 which was only 2 away. To sum it all up you want to have 100 or 200 less in the hundred's column 10 or 20 away in the ten's column and the rest go in the one’s column. An example was when I had hard numbers but made 800 first in the hundred's column 100 in the ten's column and the rest - which was 13 - I put in the one's column which made 1013-13 away.

Vihaan from Singapore American School in Singapore:

Case 1 more than 1/2 of the digits are less or = to 4 we put the biggest number and the 2nd smallest number in the 10's places.

Case 2 more than 1/2 of the numbers are 24 we put the 2 smallest numbers in the tens place and the biggest in the ones place.

The reason case one works is that the big digits and the 2nd smallest digit together make a number close to one hundred and the last two numbers make 18 at most so they don't matter.

The reason my case 2 works is that it makes the number small, but not to small and the big numbers in the tens place make the answer slightly bigger.

Subtraction, addition with more numbers

Arvin from Mountain View Academy in Canada said:

Solution (Addition 1)

First, consider the formula 10a + b + 10c + d ≈ 100, where a and b are the tens and ones digits of the first number, and c and d are the tens and ones digits of the second. This breaks the total into tens-place and ones-place contributions, helping us identify which digit pairings produce a sum close to 100.

Using the digits 1, 1, 8, and 6, list all unique tens-digit pairs: (1,1), (1,8), (1,6), and (8,6). The remaining digits automatically become the ones digits for each pairing. For example, if the tens digits are (1,8), the ones digits must be (1,6).

Now examine the tens-digit sums:
• 1 + 1 → 2_ or 3_
• 1 + 8 → 9_ or 10_
• 1 + 6 → 7_ or 8_
• 8 + 6 → 14_ or 15_

The pairing 1_ + 8_ produces a total reasonably close to 100. This is clearly the strongest candidate; the others are unlikely unless the ones digits compensate unusually well (which they do not). If two tens-digit pairings were similarly close to 100, both would need to be tested by adding the ones digits and eliminate what pair of numbers brings the number the furthest after adding the ones (regrouping if needed). For this example, the ones digits 1 and 6 give the pairs 11 + 86 and 16 + 81, both the same and equally close to 100.

Subtraction and Extras

Doing subtraction follows similar logic, but the formula changes to 10(a - c) + (b - d) ≈ 10 (10 for two digits, higher powers of 10 for more digit numbers), since the tens digits now contribute a difference rather than a sum. To obtain a result close to 10 (or whatever power of ten), the tens digits (or higher place values)** should differ by about 1, and the ones digits should be as close as possible so that b−d does not push the total far from 10. Borrowing must also be considered.

Although this method is shown for two two-digit numbers only, the same place-value logic applies to forming three three-digit numbers close to 1000: first choose hundreds digits whose sum is near 10, then use the tens and ones digits to fine-tune the total. The idea is similar, but the number of combinations grows, so more checking is required.

This heuristic aims to produce the correct answer on the first attempt by choosing the tens digits whose sum is closest to 10; only in rare tie‑cases is a second check required.

**please read the final paragraph for clarification, it talks about addition but subtraction is similar.

Syan from England said:

I decide where to put my digits by looking at the numbers I have and trying to see if they make 800 or 900 because then the tens column and one's column will add up to get you closer to 1000. If not, then put your small numbers in the hundreds column and then put your bigger numbers in the ten's columns because if your big numbers are in the hundreds, they won’t be near 800, 900 and 1000.

My strategy is to make around the 850 mark with my hundreds and then pit the tens and ones in a place. I will then add the three 3-digit numbers and move them around depending on what the three-digit numbers made. My strategy would work the best of I hit the 800 - 900 mark with large tens and ones, but it would not work if I had massive numbers in each column.

The closest I got was 1012 by having large tens and ones and small hundreds, and I moved the numbers around to get to the target a little bit closer.

Dhruv from The Glasgow Academy in Scotland said:

Operator

Sub 3 - This means they are 4-digit numbers.

Numbers

0-9

Target

Highest

Layout

Horizontal

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.

Hank from Stowe School in the UK said:

Dicey Operations

Add 1

1. Find the pair of numbers that has the closest summation to 10.

2. Make them the tens digit of the two numbers.

3. Fill the ones digits with the rest of the numbers.

Add 2

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.

Sub 1

1. Try the pair of digits that are consecutive as the tenth digit of the two numbers first.

2. Then try the pair of digits that are the same as the tenth digit of the two numbers if a pair of same number is available.

3. Then try the pair of digits with difference as 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 scenario described above is 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. Fina 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 to double-digit number 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.

The children from Twyford School, Winchester said:

Twyford School's full solution

Printable version

Roman, Elliot and Jake from Central School Te Kura Waenga o Ngāmotu in New Zealand said:

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.

Shreehari Nair from Ganit Kreeda in India said:

The strategy is like we have to keep at least two 9’s in the last row and if you get any number which is greater than 5 then keep it in first row. If you have a third 9 or 8 or a 7 then keep it in the middle row with 6 or 5. If you don’t have a 9 or 8 numbers in the 2nd row but you have bigger numbers than 5 like 5,6,7 then you must keep the first row numbers add up to 8. But if you have a 9 in the middle row then the first row add up should be 8.

Elli from Shatin Junior School ESF in Hong Kong said:

Elli's full solution

Sofia from Shatin Junior School ESF in Hong Kong said:

Sofia's full solution

The children from St. Martin's Catholic Primary in the UK had lots of ideas.

Moses said:

I used the ones to make tens and I used tens to make a hundred. I thought about placing the biggest numbers added to the smallest numbers to not make it too big.

Myla and Otto said:

We made a column method in our head and we helped each other. We tried to make the ones add up to 10 and the tens add up to 90.

The students from William Konkin Elementary School in Canada had lots of ideas. Aubree and Sienna explained:

Oscar and Nelson explained:

Christopher Wang from King George V School in Hong Kong said:

First of all, suppose we have a 2x2 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.

Christopher's full solution

We also received similar solutions from: Marvin from St. Mary's International School in Japan; Ayah from Doha college in Qatar; Advait from Kings' School Al Barsha in the UAE; Shivaprasad from The GYM Foundation in India. Thank you all for sharing your ideas with us.