For this problem, lots of people submitted their methods for finding the times table and the shift, and some people sent in answers to the questions in the problem.

Here are your methods for level 1 and level 2, followed by your answers to the questions, followed by your methods for level 3 and level 4.

Van Anh from British Vietnamese International School in Vietnam sent in a useful insight for finding the times table and shift at level 1 and level 2:

In the times table, even if the times table is shifted, the distance between the two numbers will be the same. Let's say our 2 times table, which has a distance of 2, when shifted by 101, will become 103,105,... which still has a distance of 2. You can use this to help when come dealing with these kinds of problems.

Roshni from Tanglin Trust School in Singapore, EDSR from Priestlands School in the UK and Oren from Sunnynook Primary School in New Zealand used the same method for finding the times table and shift at levels 1 and 2. This is Roshni's work:

In this level, there is a common difference(d) between consecutive terms (this is the distance that Van Anh was talking about). To work out the common difference(d), you simply find the difference between any two adjacent numbers. For example in the sequence 14,22,30,38,46, I could work out the common difference by doing: 22-14 or 30-22 or 38-30 or 46-38, all of which give the same answer of 8. Once you have found the common difference, you can then subtract (common difference $\times$ n), so you are subtracting 8,16,24,32,40 from the sequence which results in 6,6,6,6,6

Therefore, the shift is up 6. If the difference between the (common difference $\times$ n) and the sequence is negative, then it is a shift down.

Sequence:                          14, 22, 30, 38, 46

Common difference $\times$ n:       8, 16, 24, 32, 40

[Result of] subtraction:          6,  6,  6,   6,  6

So, for the example used above the table is 8, it is shifted up by 6.

What can you say if the numbers are all odd?

Anh Minh from British Vietnamese International School Hanoi said: the times table you are looking for will always be 2, it will shift up/down 1.

Anh Minh is right that shifting the 2 times table (all even numbers) up or down by 1 will give only odd numbers, but this is not the only way that the numbers could all be odd. 

Alex said: That means all the numbers are from an even number's time table and that the shift is an odd number. This is because any number times any even number will always give an even number, and when even numbers are shifted by an odd number they always give an odd number.

What about if they are all even?

The Marshmallows and EDSR correctly said: the multiplication table is bound to be even.

John and Jack from Trinity Grammar in Australia gave a fuller explanation: It must be an even number [times table] shifted up by an even number.

Or a mixture of odd and even?

Alex said: This means the time table is an odd number as they're the only ones that alternate between odd and even numbers. The shift can be anything because it may turn odds into evens and viceversa but there will still be both odds and evens.

What can you say if the units digits are all identical?

The Marshmallows said: the times table is bound to end in zero, which Alex and John and Jack described as a multiple of ten.

What if there are only two different units digits?

EDSR said: The table's units digit will have to be a five, which Alex and John and Jack described as a multiple of five - although it must be a multiple of five but not a multiple of ten.

What can you say if the difference between two numbers is prime?

EDSR identified that: The prime number will be the multiplication table.

Alex was more precise: Those 2 numbers are consecutive terms of a prime number's time table. The shift can be anything as it won't change the difference.

What can you say if the difference between two numbers is composite (not prime)?

EDSR and The Marshmallows both explained the same idea. The Marshmallows said: The times table could be composite but it could have numerous times tables i.e. 4 times table is also in the 2 times table.

Again, Alex was more precise: The numbers are either multiples of a non-prime times table or non-consecutive terms of a prime times table.

Can you explain how you worked out the table and shift each time, and why your method will always work?

The Marshmallows explained the steps which form the foundations of all of the methods which were sent in:

First you must find the difference between all the numbers. If they are all the same then you know the times table then add (or subtract) to the original table then you have the shifted answer.

If the difference between the numbers is not the same throughout, then you need to find a common times table before working out the shift.

How can you find the common times table?

Hayley from Tarremah Steiner School in Australia, Charlotte from Ryde School in the UK and EDSR all assumed that the common times table would be the smallest of the differences. This is Hayley's example:

My numbers were: 

180   438   481   610   653   

First you have to find out how much there is between each number, for instance, the difference between 481 & 438 is 43 & the difference between 653 & 610 is also 43.

Then you get the lowest number you have (180) and divide it by your difference (43)

180$\div$43 is approximately 4, then you go 4 $\times$ 43 = 172, then you take 172 from 180 which is 8

So you end up with your times table (43) and the number that you've shifted it by (8). 

It is almost like a linear equation $y=mx+c$ . 

So you have to find your multiple (43) and your constant (8) so you end up with $y=43x+8$

Charlotte gave the steps of this method in more detail. Click here to see Charlotte's work.

Alex explained why this method doesn't always work:

The highest common factor of the differences will be the times table. At first I thought it would just be smallest difference but then I came across this:

given that the 5 numbers are 9 18 24 30 39

the differences would be 9 6 6 9

so you could think the time table is 6, but it's actually the HCF of 6 and 9, which is 3. This is true because using the 6 times table you can go from 18 to 24, but not from 9 to 18. So the largest time table that will satisfy both will be 3.

Ms Rusnock from Stuart Country Day School of the Sacred Heart in the USA, Roshni and Oren used a more robust method. This is Roshni's work:

You should first reorder the sequence given to you to increasing order. For example, you are given 128,213,60,264,179. The new version would be 60,128,179,213,264.

First, work out the difference between any two numbers next to each other - for example do 179-128 which is 51.

Then work out the difference between another pair of numbers that are next to each other (one of the numbers may have been used for the previous subtraction), so for example do 213 and 179 which is 34.

Noticeably, the differences are different - 51 and 34. So you should find the Highest Common Factor of [all of] the differences, which is 17 in this case, which will give you the table.

Then, for each number in the sequence, find the closest [multiple of 17] that is less than that number. So for the example it is 51, 119, 170, 204, 255. Then, calculate the difference between the closest [multiple of 17] and the ordered sequence which is 9,9,9,9,9.

Sequence:                        129,213,60, 264,179

Ordered Sequence(OS):     60, 128,179,213,264

Difference:                           68,  51,  34,  51

Highest Common factor for difference: 17

Table closest to OS:            51, 119,170,204,255

Diff between tab and OS:     9,    9,    9,   9,    9

So, for the example used above the table is 17, shifted up by 9

Alex wrote a program to find the times table and the shift. This is Alex's description of the program:

I wrote a python script to which you input all 5 numbers one at a time followed by an enter, they are all put into a list which is then sorted by ascending order. Then the difference between each pair of 2 consecutive terms is then calculated (So the difference between the 1st and 2nd, 2nd and 3rd, 3rd and 4th, etc) listed and sorted in the same way as previously.

The highest common factor of the differences will be the times table. I work out the HCF using the Euclidean algorithm, which takes in only 2 values at a time, however we need to calculate the HCF of the whole thing, to do so we use the following property:

HCF(w,x,y,z) = HCF(HCF(w,x),HCF(y,z))

So this way it arrives at a single term which is the times table.

I used the Euclidean algorithm for the HCF as it makes the code shorter and more efficient, but it could also be computed with a classical list of factors and then picking the largest one.

The shift is calculated using the MOD function (the remainder of a division). When any of the numbers is divided by the times table the remainder will be the shift upwards.

So if the number is 57 and the times table is 20, so MOD(57, 20)= 17 (in python MOD is written as %) so here the shift would be 17 upwards.

The times table and the shift are then outputted.

Click here to see Alex's code, or here to see it as a PDF.

Lots of people added together three numbers and noticed something special! For example, Emma, from Interlakes Elementary, told us her findings:

I figured out that, if you kept adding the numbers together, they will always come to a multiple of 3 every time you do it.

Charlie, from St. Cecilia's Wandsworth, noticed:

What we did was: we started with 3 numbers, then we added them together, and we noticed that some were odd and some were even but they were all multiples of 3. Then we tried adding 4 numbers, and we found that the answers were 1 more than the numbers in the 3 times table. Choosing 5 numbers we got answers which were 2 more than the 3 times table. We guessed that adding 6 numbers would give answers back in the 3 times table.

Great observations! Bethan, Gareth and Aditya, from St. Nicolas CE Junior School, offered an explanation of this. Bethan wrote:

When you have your three numbers, say 4, 7 and 1, each of these numbers is 1 more than a multiple of 3.

So 3+1, 6+1 and 0+1.

Then when you add them together, you can add the 3, 6 and 0 together which makes a multiple of 3, plus the three 1s left over will add together to also make a multiple of 3.

This makes your overall answer a multiple of 3.

Good! (Is this related to either Charlie's or Alison's method?) Aditya used the same method to notice:

Adding together 99 numbers would give a multiple of 3, and 100 numbers would equal a multiple of 3 plus 1.

Brandyn from Garden International School considered what happened when he added sets of 3, 4, 5, 6 and 99 numbers from the bags below:

 

 

Image

 

 

Choosing four numbers from the bags above gave the following totals, all 1 more than (or 2 less than) multiples of 3:

 

 

Excellent. Thanks!

Miraya from Heckmondwike Grammar School in the UK wrote:

According to all numbers I have tried, I think it is not possible to find a set of five numbers that do not have a set of three that sum up to a multiple of three.

Samuel from Seely Primary School in England, Jimin (James) from BSKL in Malaysia, Harvey from St Nicholas in the UK, Ekaansh from Dulwich College Beijing in China, Kai and Wesley from Wilson's School in England, Ariel from Diocesan Girls' School in Hong Kong, Aditya from Bishop Vesey's Grammar School in the UK all used the idea of remainders when dividing by 3. Ekaansh wrote:

I do not think this is possible. This is because any integers we pick would be one of these:

• Exactly divisible by 3 e.g. 3,6,9 etc.
• Multiple of 3 plus remainder 1 e.g. 4, 7, 10 etc.
• Multiple of 3 plus remainder 2 e.g. 5, 8, 11 etc.

Harvey represented the numbers in a different way, but the idea the same:

You can have three groups of numbers: multiples of three minus 1, multiples of three, and multiples of three plus one. If you continuously add or subtract three you can get one of three numbers: 1, 2, or 3.

Jimin (James) continued:

We only have to add up these [remainders] (1-2-0) because if they add up and make a number which is a multiple of 3, the left over 3s in the number will make a multiple of 3 anyways. 

Samuel explained this idea using letters for the types of number:

I represented the numbers that are one more than a multiple of three as $x$, numbers that are 2 more than a multiple of 3 as $y$, and numbers that are a multiple of 3 as $z$.

$x$+$y$+$z$= a multiple of 3.

Example: 1+2+3=6

$z$+$z$+$z$=a multiple of 3

Example: 3+3+3=9

$y$+$y$+$y$=a multiple of 3

Example: 2+2+2=6

$x$+$x$+$x$=a multiple of 3

Example: 1+1+1=3

Since every number has to be either $x$, $y$ or $z$, and that these combinations work, it is impossible to produce a set of 5 whole numbers that any 3 of them do not add up to a multiple of 3.

This is because in a group of 5, it is not possible to have no groups of 3 [of the same type], whilst also not having one of each type.

This is Ariel's explanation of why there will always be three numbers that add up to a multiple of 3:

First, if there is at least 3 of a kind, there must be 3 integers that add up to a multiple of 3 because any 3 of a kind sums up to a multiple of 3.

Secondly, if there is no 3 of a kind, the combination of the remainders must be 2, 2, 1, in any order. In that case, there must be 3 integers that add up to a multiple of 3 since 0+1+2 sums up to a multiple of 3.

Therefore, there must be 3 integers that add up to a multiple of 3 no matter what 5 integers you choose.

To be absolutely sure that it's impossible, Aditya found all of the possible sets of remainders that 5 numbers could have:

Image

This idea of using remainders is very useful in number theory and is called modular arithmetic. 

Emily from Garden International School in Malaysia, Rin from Wren Academy in England, Saanvi from UK, Mahdi from Mahatma Gandhi International School in India, Sanaa from Heckmondwike Grammar School in England and Sunhari all used the same idea, but expressed the numbers using algebra. This is Emily's work:

[All] integer[s] can be divided in three groups which are $3n,$ $3n+1$ or $3n+2$. $n$ can be any whole number. 

We have 2 [ways of making a multiple of 3]

The 1st option is we have at least one number from each group. We can use the letters $n,$ $m$ and $l$ to show the different numbers. So when we add up the numbers it'll look something like this, $$3n+(3m+1)+(3l+2)= 3(n+m+l)+3= 3(n+m+l+1)$$As you can see this number is definitely a multiple of $3$ because we can look at the

numbers in the bracket as [an] integer and $3$ has to multiply that number. Therefore $3(n+m+l+1)$ can be divided by $3$.

Note that $3n, 3n+1$ and $3n+2$ represent a multiple of $3$, the number $1$ more than that multiple of $3$ and the number $2$ more than that multiple of $3$. That is why it is important to use $m$ and $l$ as well as $n$.

The second option is we have at least three numbers from one group. 

Sunhari showed this option algebraically:

$(3t+1) + (3j+1) + (3w+1)$

$= 3t + 1 + 3j + 1 + 3w + 1$

$= 3(t+j+w) + 3$

Therefore, it would be a multiple of $3$.

$(3x+2) + (3g+2) + (3d+2)$

$= 3x + 2 + 3g + 2 + 3d + 2$

$= 3(x+g+d) + 6$

Therefore, it would be a multiple of $3$.

Sanaa and Rin explained carefully why this means that there will always be three numbers that add up to a multiple of three by systematically considering different cases. Here's Rin's explanation:

1. If there are three or more multiples of $3$ within the five given numbers, we can add those to make a multiple of $3.$ 
   
    
2. If there are one or two multiples of $3$, other three numbers must be either $3n+1$ or $3n+2.$ If the remaining numbers include:
   1. Three or more $3n+1$ s
      
      We can add these three to get a multiple of 3.
   2. Three or more $3n+1$ s
      
      Similarly, we can add these to get a multiple of 3.
   3. Otherwise
      
      We can pick one each from groups of $3n, 3n+1, 3n+2$ and add together.
      
       
3. If there is no multiple of 3, the five numbers can be written as either $3n+1$ or $3n+2.$
   
   For any combination of five numbers, three or more must be in the form of $3n+1$ or $3n+2.$
   
   Adding numbers in the same group would give a multiple of $3.$ 

Therefore for every combination of five numbers, there are three numbers that add to give a multiple of $3.$

Saanvi and Mahdi used the same idea, but with $-1$ instead of $2$. Everything else works out the same. Here is some of Mahdi's work.

Image

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Shaunak from India, Roy from Dulwich College Beijing in China, Ci Hui from Queensland Academy for Science Mathematics and Technology in Australia, and Ruoshui from UK gave a similar argument to those above, but used notation or language of modular arithmetic. You can read Shaunak's argument here and Roy's here. Ci Hui also included a table showing different options for the three numbers chosen depending on the remainders. 

After explaining the case for 5 numbers using similar ideas to those above, Ruoshui then gave the following generalisation: 

The problem generalises to $2n - 1$ integers, out of which the sum of $n$ of them is divisible by $n$.

What do you think about this generalisation? Perhaps you could test it using different numbers. Which ideas used above might help you to explore this generalisation?

Jenny from Tapton Secondary School related these arguments with remainders to another principle that can be useful in mathematics: the pigeonhole principle. This says that if you have $n$ pigeon holes and at least $n+1$ pigeons, at least one pigeon hole must contain two or more pigeons. As Jenny notes for this problem, 

It works because of the pigeonhole principle. When you divide any integer by three, the remainder is 0, 1, or 2. So, with five numbers, there must be at least three numbers in there that, when their remainders are added, total to a multiple of three.

Can you connect Jenny's explanations with the arguments above?

Thank you to everyone who sent in solutions to this problem!

Henry, from St. Hugh's, answered our question at the end:

If the bags contained 3s, 7s, 11s and 15s, can you describe a quick way to check whether it is possible to choose 30 numbers that will add up to 412?

He said the following:

It is impossible to make 412.

The starting number is 3 and the difference between the numbers is 4.

If I choose 30 numbers and add them all up, I will get a number that is 30x3=90 more than a multiple of 4.

But, 90÷4=22 remainder 2, so I will get a number that is 2 more than a multiple of 4.

But since 412 is a multiple of 4 (not 2 more than a multiple of 4), it won't work.

Well spotted! Luke, from Cottenham Village College, said this in a more algebraic way, using a tool called 'modular arithmetic':

All of our numbers are of the form 4x-1, for x=1, 2, 3 or 4. Therefore the sum of 30 of them comes to $4(x_1+x_2+\dots +x_{30})-30$, which is 2 mod 4 (i.e. 2 more than a multiple of 4). But 412 is 0 mod 4 (i.e. 0 more than a multiple of 4), so this sum cannot be equal to 412.

If you are unfamiliar with Modular Arithmetic, you might like to take a look at this introductory article.

Luke also gave his thoughts on the interactivity:

The numbers in the bag always form part of a linear arithmetic sequence, and so the number in the x-th bag is mx+c. Consecutive numbers are always a fixed distance m apart. This means that we can read off the value of m easily, and then find the value of c. We can then conclude that, if you choose z numbers from the bags, their sum will be of the form $m(x_1+x_2+\dots +x_z) + cz$, as all the numbers are of the form mx+c, for different values of x. This is obviously cz more than a multiple of m.

To illustrate this method with an example, take the numbers in the bag to be 1, 4, 7 and 10, as in the original example, with z=4.

Their differences are all a multiple of 3, so by analysing them mod 3 (i.e. by looking at their remainders when they are divided by 3) we find that they are all of the form 3x+1.

As z=4 and c=1, they must be cz=4 more than a multiple of 3.

Since 4 is 1 mod 3 (i.e. dividing 4 by 3 gives remainder 1), 4 numbers selected from the bags must add to give 1 more than a multiple of 3.

Nice!

 

Vinay from Ripley Valley State Secondary College in Australia, Leala from Notredame in the UK, Tyler from Burnside Primary School in Australia and R from the UK worked out the rules for specific versions of the game. Every time you play, the rules are different, so people's answers are different. Here are some examples of the sets of rules they found.

Tyler:

Blue light turned on by repeating pattern of adding 2 starting from 1 so the pattern is 1,3,5,7,9,11 and so on. 

Yellow is every 3 starting from 2, so 2, 5 8, 11, 14, 17,20 etc.

Green is every 9 starting from 2, so 2, 11, 20, 29, 38, 47 etc

Red is every 8 starting from 5, so 5, 13, 21, 29, 37, 45 etc.

Leala:

Yellow- multiples of 3

Red- going up in 8 them minus 2 (8$n$-2)

Blue- going up in 8 then minus 5 (8$n$-5)

Green- going up in 6 and then minus 2 (6$n$-2)

Shaunak from Ganit Manthan, Vicharvatika in India explained how to find the rules. You can read Shaunak's method below, or watch Shaunak's video by clicking here.

Here is a strategy to work out the rules controlling the lights:

1: Find the rules for the Yellow light - Start by keeping zero in the box. Keep on increasing the number in the box by one. Note down the first 5 numbers that satisfy the rule for the Yellow light.

Next, find the difference between any two consecutive numbers in this list. Call this difference $n.$

Then, find the remainder obtained by dividing any number from this list [by $n$]. Call this number $b.$

The rule is - Numbers $b$ more than [a multiple of] $n.$

So, any number that can be expressed as $a\times n + b,$ where $n$ and $b$ are those computed above, and a is any number, satisfies the condition for the Yellow light.

In a similar manner, one can find the rules for the Red, Blue and Green lights. These three also follow the $a\times n + b$ property.

Example:

I started with 0. The first five numbers that satisfy the rule for the Yellow light are - 2, 5, 8, 11, and 14.

Now, I need to find $n,$ which is the difference between two consecutive numbers in this list. 2 and 5 are consecutive numbers in this list, so 

$n$ = 5 - 2 = 3

Next, I should find $b$, which is the remainder obtained by dividing $n$ by any number in this list. Because 11 is a number in the list, and $n$ is 3, so the remainder obtained is 2 (from 11$\div$3).

The rule is - Numbers 2 more than multiples of 3.

So, any number that can be expressed as 3$a$ + 2, satisfies the above rule.

If $a$ is 343, I will get a number larger than 999, or a 4-digit number.

Let me put $a$ as 343 and see if the Yellow light gets switched on...Yes! It worked! 

Using the same strategy one can find the rules and properties for Red, Blue and Green lights.

 

Students at Clifton Hill Primary in Australia worked hard to beat the computer at Got It with a target of 23, and with numbers 1, 2, 3 and 4 to choose from.

J and C said:

You can win the game with a target number of 23 and range numbers of 1-4 by trying as hard as possible to land on the number 18. By landing on 18, you make your opponent off by 1 of getting the end number. If you get the number 18, your opponent will be stuck and you have a guaranteed win.

Danica and Eve took this a little bit further:

Can you find a winning strategy?

Yes, you can by getting to 13 then the other player has to say their number (1-4) you can get to 18.

After that, no matter what number they say there will always be a way for you to get to 23.

Leon and Jack built on this even further:

There are three key numbers:

8

13

18

Firstly, you have to be the one that gets the total to eight. You can do this by starting with 3 and the opponent has to let you get to 8. Once you have accomplished that, it is their turn with a total of 8. They then can't stop you from getting to 13. After you get 13 they can't stop you from getting to 18. After this, they can't stop you from reaching 23.

So, in fact rather than there being three ‘key numbers’, it sounds like you’re saying there are four key numbers: 3, 8, 13, 18.

Golden Eagles Class at Ditchingham Primary Academy described their strategy in a similar way:

We found that it was easier to win once we had found the stepping stones for the target number.  This was a key number that if we reached we could win any game.

As we could add up to 4 each time the difference between each number was 5. The stepping stones were 3, 8, 13, 18 for 23.

For example, if we started with 3, our opponent could only make it up to 7, we could then add 1 to make 8. The most they could then make is 12 but with all variation of numbers we could then make 13 and so on up to 23.

When we played the computer using this strategy and we went first we always won, however if the computer started we always lost. It was possible to beat an opponent if they went first, but only if they hadn't worked the stepping stones out. Once they had we always drew.

We tried using 17 as a target number and found stepping stones at 2, 7, 12 using the numbers 1-4. This was again as the difference between these numbers was 5.

We would like to try dropping the 4, using numbers 1-3 and seeing if the stepping stones reduced to a difference of 4.

I like the fact that you are making a conjecture about how you could apply your strategy to a version of the game using only numbers 1, 2 and 3, Golden Eagles Class.  This is exactly what mathematicians do!

This idea of ‘key numbers’ of ‘stepping stones’ is crucial in this game. Mateusz and Joseph from St Ann’s Primary agreed with this approach, so did Amelie and Sam from St Francis of Assisi Primary School; Gabriel from Yavneh College; Alaric from Haydon Wick Primary; Harry from St Michael’s Preparatory School; Ruby, Alice, Meiyla, Lilly, Olivia and Sophie from Anston Greenland’s Primary; Joy, Akosua, Asuo and Kumi from Torrens Valley Christian School in Australia; as well as three groups of students from Ganit Kreeda, Vicharvatika, India.

Anirudh from Hamilton Primary School shared this video, in which he explains how he worked backwards to find the stepping stones:

And in this video, Kaana from Ganit Kreeda explains the importance of multiples of 5:

Can you see how Anirudh’s and Kaana’s solutions are connected?

Joseph from Jesmond Park Academy sent a very comprehensive solution:

I initially approached this problem by just playing some games with the computer and tried to recognise some patterns. I started by playing first but then swapped to the computer playing first. It was here that I noticed that it would always start with the same first number depending on the target number.

I noted this down and also noted that it would try to reach certain values consistently. I called these 'landing values' and wrote them down.

From here, I noticed that there was a fixed distance between each landing value that was 1 greater than the largest number I could choose. This revelation led to further investigations that led to the process below.

For anyone who wanted to consider solving this problem, or ones like it, I think the best first step is to just play around and consider different values. After you've considered different values, you can assess whether there is a pattern or not. This pattern is the real starting off point for finding a general solution.

Thank you, Joseph, what a lovely description of the process you went through. We agree that ‘playing around’ with a new problem is a great way to start. The exploration then leads to some noticing, as Joseph said, and then some more systematic working, some generalising and even proof.

Here is Joseph’s general strategy for winning at Got It:

The numbers that are chosen to be added up range from 1 -> n, where n is a positive integer and the spaces between 1 and "n" are integers. (In fact I wonder whether they have to be consecutive numbers, Joseph...)

For example, "1, 2, 3, 4" where n is 4.

The player will have a target number, which we will call "m". To begin, player 1 will write down the target and subtract (n+1) away from the number "m" until there is a number that they are able to choose, we will call this "b".

For example, consider the target is 22 (so m=22) and the player can add up to 4 (so n=4), they will subtract 5 (n+1) until they reach the number 2 (so b=2).

Player 1 will choose this number and begin the game. From this point forward, it is Player 1's objective to reach the next multiple of (n+1) that starts from "b". In our example, these "Landing Values" would be 7, 12, 17 and finally, 22. Player 1 will always be able to force a Landing Value on their turn assuming they started from "b" and their opponent plays perfectly.

If player 1 aims for these numbers, Player 2's choice will not matter as they will end up unable to reach the target number.

The exception to this trend comes if the target number is a multiple of (n+1). This is because you can't keep subtracting (n+1) to reach the initial number "b" as you will eventually reach (n+1) and the next number would be 0.

For example, if we could choose to add up numbers between 1 and 4, if the target was 10, we could subtract 5 (n+1) once.

Following this, subtracting 5 again would reach 0, which isn't an option with our initial conditions.

Thus, the first player would have to choose any number between 1 and 4.

Assuming that there is a perfect opponent, they would reach the multiple of (n+1) instead, in this example it would be 5. The game would continue as stated previously, resulting in player 2 winning instead.

Thus, you can reach the conclusion that it is only sensible to be Player 1 if the target number is not a multiple of (n+1). If it is, then you should choose to be Player 2.

Abishai expressed a general solution in a very similar way, so did Xavier, and so did students from Ganit Kreeda, India. Thank you to you all. How fantastic that we have used our mathematics to find a winning strategy that will ALWAYS work when playing Got It.