More less is more

Thank you to everybody who sent us their ideas about this problem. We'll be focusing here on the solutions which explained a method for finding the highest possible score for the four versions of this game.

Sum-sum

The Golden Eagles and Snowy Owls from Anston Greenlands Primary School began by looking at the tens digits. They explained:

You need to put the largest digits in the tens position.

63 + 55 < 63 + 61

Matilda from Stanborough School in England had a similar strategy, and explained which of the largest digits should go in which tens positions:

You put the biggest digit on the right side (in the first slot) then "balance" it out by putting the second biggest on the left side (in the first slot). E.g. if you had the digits 2 2 2 3 3 3 4 5 you would put the 5 on the right side and then "balance" it out by putting the 4 on the other side. Then you would put the next biggest number on the left side in the 3rd slot.

Inaya from Ganit Kreeda in Vicharvatika, India thought about this in terms of trying to make the tens digit addition equal on both sides:

Take the 4 largest numbers.

Place them in the tens place to make the tens addition of both sides same (situation 1) but if it is not possible then make the right’s tens addition side bigger (situation 2).

(For situation 1) Take the rest of the numbers and make the bigger ones addition on the right ones side and the smaller ones addition on the left ones side but keep the additions difference as small as possible.

(For situation 2) Take the two biggest numbers and place them on the left ones side to make the tens addition as big as possible and then place the rest of the numbers on the right ones side.

Advaya from Ganit Kreeda described this strategy using three simple rules:

Rule 1: Inequality should be right.

Rule 2: Tens place of LHS and RHS should be close or equal.

Rule 3: If tens place equal, greater ones place sum should be on RHS.

Noah from St Augustine's Catholic Primary School in the UK worked out the largest total when the numbers given are 1-8:

(B=blank.) I started with using the 7 and the 8 to make 7B + BB < 8B + BB. This maximises our score while keeping the inequality true. Then, I added a 6 to the left side, not a 5. This makes sure that our score is at least 130. Now we have 1, 2, 3, and 4 left. We can use this to make the solution 71 + 63 < 84 + 52. This gives us a score of 134.

Siddharth from Ganit Kreeda explained why this is the highest possible score:

With the numbers 1 to 8, the highest possible score is:

83+51<74+62

The answer is valid because the left side equals 134 and the right side equals 136, making a valid inequality. 135 could be possible, but then both answers would be equal, thus making them invalid.

Take-take

The Golden Eagles and Snowy Owls explained how to maximise the score in this version:

66 – 45 < 66 – 35

In take-take, put the largest two digit numbers on the left hand side of each calculation and the smallest possible numbers after the subtraction sign.

Aarav from KHSB in India described a more detailed method for maximising the score:

A B - C D < E F - G H

First take the largest 2 numbers.

Possibility 1: They are equal. Place them in boxes A and E.

Possibility 2: They are not equal. Place the greater one in box E and the smaller one in box A.

Now, place the smallest 2 numbers in boxes C and G so that either the answer you get when you subtract the right side tens is more than the left side tens or both the sides' tens are equal. Finally we again have 2 possibilities for the remaining numbers.

Possibility 1: The tens of both sides are equal.

Place the remaining numbers in boxes B, D, G and H so that the right side ones are more than the left side ones.

Possibility 2: The tens of the right side is more than the left side tens.

Place the remaining numbers in boxes B, D, G and H so that the left side ones are more than the right side ones.

Aarav has made sure that the inequality will be correct, but is there anything else we need to think about here to make sure we get the largest possible score?

Hrishikesh from Ganit Kreeda had some ideas about how to get the largest possible score in this version:

1. Place the 2 highest numbers in the tens place of both minuends (the minuends are the numbers we are subtracting from). If the numbers are unequal, place the higher number on the RHS. The ones places of the RHS and LHS minuends get the 3rd and 4th highest numbers.
2. Place the 2 smallest numbers in the tens places of both subtrahends (the subtrahends are the numbers being subtracted). Try to equalise the LHS and RHS after subtraction of the tens place. If this is not possible, minimise the difference between the LHS and RHS tens places after subtraction, but make sure the RHS is greater than the LHS.
3. Place the remaining 2 numbers so that you get the highest possible score by minimising the difference between LHS and RHS after subtraction.

I agree that putting the two largest numbers in the tens places of both minuends and putting the 3rd and 4th largest numbers in the ones places of the minuends will always give the largest possible score. But deciding where exactly to put those numbers (and the four smaller numbers) seems to depend on the numbers given. Using Aarav's method of naming the cells (A B - C D < E F - G H), can you find examples where it's better to put the largest number in place A, instead of place E?

Take-sum

The Golden Eagles and Snowy Owls began to think about how to get the largest possible score for this version of the game:

In take-sum, you need to put the largest two numbers on the far left hand side.

66 – 23 < 55 + 64

It certainly seems like making the first two-digit number as large as possible is going to be a helpful strategy here. Dhruv from The Glasgow Academy in Scotland, UK followed this up by subtracting the smallest possible number from the largest:

The numbers I was given were: 6, 6, 4, 4, 4, 1, 1, 3

The equation I made was: 66-11<44+43

2nd Time

6, 4, 1, 6, 1, 6, 2, 1

66-11<64+21

Strategy: Put the biggest number and the smallest number on the left-hand side and the rest of the equation follows on.

Advaya from Ganit Kreeda continued this strategy by making the numbers on the right-hand side as large as possible:

Rule 1: Highest numerals should subtract lowest numerals. E.g. numbers: 1,3,9,8,6,3,7,4.

98-13<a d + x y

Rule 2: Once LHS done, do high sum on RHS. E.g. 74+63

Aarav from KHSB recognised that by using this strategy, the inequality might end up not being true, so some tweaks might need to be made:

A B - C D < E F + G H

Create the highest possible 2-digit number which can be made from the given digits and place it in boxes A-B. Then, create the smallest 1 or 2 digit number and place it in boxes C-D. Lastly, place the remaining numbers in boxes E, F, G and H so that their sum is more than A B - C D. In case this does not work try decreasing A B and increasing E, F, G and H until it works.

Hrishikesh from Ganit Kreeda had an idea of a tweak that could be made in this case:

If the inequality does not stand true, exchange one of the RHS tens places with the LHS minuend ones place.

This seems like a good place to start making adjustments, as changing that ones place won't reduce the score by very much!

Sum-take

Dhruv from The Glasgow Academy found that in this version, it often isn't possible to satisfy the inequality:

The answer usually is impossible but if is possible then put the biggest and the second smallest on the right and put the middle numbers on the left.

Hrishikesh from Ganit Kreeda agreed with this method, and also had some suggestions for how to make adjustments if the inequality isn't satisfied:

1. Place the 2 highest numbers on the RHS minuend. The higher number goes on the tens place, and the lower one goes on the ones place.
2. Place the 2 lowest numbers on the RHS subtrahend. The lower one will go on the tens place, and the higher one will go on the ones place.
3. With the remaining 4 numbers, make the LHS as high as possible while making sure that the inequality stands true.
4. If the inequality doesn't stand true, exchange the lowest LHS tens place with the RHS subtrahend ones place. If they both are equal, it is impossible to satisfy the inequality.
5. If, even after you do that, the inequality does not stand true, take the higher LHS tens place and exchange with the current RHS subtrahend ones place.
6. If the inequality still does not stand true, it is impossible to satisfy the inequality.

Tip: after following an algorithm, check if you can exchange the places of a few numbers while making the inequality stand true before submitting your answers.

For the version with the numbers from 1-8, this method doesn't quite give the largest possible score. It's very close, though!

Kanaa and Avic from Ganit Kreeda found the largest possible scores for all four versions of the game using the numbers from 1-8. Take a look at Avic's full solution to see how these games can be solved using a combination of logic and trial and improvement.

The following students also sent in similar solutions to this problem: Nikhil, Arham, Warren and Musab from Doha College in Qatar; Eva, Sara and Natalie from Nord Anglia International School Abu Dhabi, United Arab Emirates; and Vishnu, Shivashree, Reyansh, Nehtanya, Arjun and Avyansh from Ganit Kreeda. Thank you all for sharing your ideas with us.