I’ve no problems in math in any placements: schools, universities, but I always been struggled at the olympiads. This questions wants some impossible thinking during short time, which requires that you knows these types of questions and know how they are solving, otherwise you’re just a genius, who can solve any sort of problems in few hours
That’s because they’re used to test the best young mathematicians in the world these problems aren’t meant to be easy and anyone who says they are is probably lying
@@ThoseInterestingStories i mean like the only hard problems are millennium problems and most of the time they do not even have a right answer let alone a proper solving technique known to current principles of mathematics
Fun fact: The person who solved the problem entirely (got 10/10) is Romanian and he also got a max score in the IMO that year. Being Romanian too, I see this guy as the Romanian Jesus of Mathematics, you can show him any problem and he will solve it for. Anyway, completely coincidentally, the author of the problem is also Romanian, and the original text used "n natural number" instead of "2023", but they changed it before the contest as it probably would have been an overkill.
Currently at 7:08 and there is one more constraint on k. If k = 1, Alice can choose 1 or 2 for that one number, and Bob’s task is impossible. So 1 < k < 593
5:00 I solved this a different way: if there are more than 50 numbers in the subset, we know that they can't all be odd / can't all be even (since there are only 50 odds and 50 evens in the original set) - thus, there is at least one odd number and one even number, which we can add to get an odd number.
I feel like this is the natural way of doing it (I know I did). Also interestingly this is actually the same method. However, the lessons learned from the smaller problem stated in the video are from turning your proof into one of pairs and pigeonhole. But I wouldn’t have got the same “tooling” from your (and my own) explanation. Interesting how different interpretations lead to different progressions.
My thoughts after hearing the problem: The logical limit would be bΔ = 2023Δ/2, where b = 2023 - k, giving k = 592 (rounded down) If Alice takes the 592 largest numbers, Bob can take the rest (excluding numbers summing to 1916, to make it equal). If Alice takes the 593 largest numbers, the sum is greater than the sum of the remaining numbers, so Bob cannot make an equal sum. Intuition says that there will always be plenty of small numbers for Bob to choose from to make the numbers equal, but I'll try to come up with a logical method: Once Alice chooses her numbers, Bob chooses the remaining numbers from the largest down, until his sum exceeds Alice's sum. He then chooses one extra number, that is the negative of his excess (for example, if his chosen numbers go down to 61, and his sum is now 25 greater than Alice's sum, he adds the number -25 to his set, to even them out). He then reduces the smallest non-negative number to the next available number (e.g. 61 => 60), and then increases the negative number by the difference to keep it even. He then sends the new gap (now at 61 in the example) bubbling up his set, so he can increase the negative number into the positives, until it eventually meets the rest of his set (e.g. [-24, 60, 62, 63] => [-23, 60, 61, 63] => [-22, 60, 61, 62] => [-21, 59, 61, 62] etc). In the case where Alice chooses the largest 592 numbers, there are ~60 legal spots for the floating number to end in. To prevent this method from working, Alice would have to have to occupy at least 60 of the first ~120 numbers (blocking spots where either the floating number or the bubble would want to be in each iteration). However, doing this would increase the number of Bob's unused numbers faster than she could block them (e.g. if she chooses the largest 591 numbers and then also 120, it would reduce the sum by ~1300, meaning Bob would only have to choose numbers down to ~80, so Alice would now have to block 80 of the first ~160 numbers).
This is a great problem and some interesting solutions, but you jump from one idea to another so quickly and oftentimes briefly mention relevant information or use so much previously defined variables without reminding the viewer what they were, that closer to the end of the video it became unwatchable.
The way the problem is presented confused me at first since it didn’t say any k of the set just exactly k. Maybe it’s a standard that these problems are worst case scenarios but I assumed it was alice working to maximize k, in which case she would choose the smaller numbers over the larger.
Generally speaking, olympiad problems give you the absolute least amount of information to be 100% certain what the question is. If the question says "find the greatest integer k such that..." Then they do just mean the result holds at exactly k. If they wanted it to hold for all values up to k they'd say "find the greatest k such that for all m ≤ k" the result holds.
Not sure what it is yet, but can't be more than 592. Otherwise Alice would pick the highest numbers and Bob would not be able to get a sum of equal size.
To my own surprise I actually solved this one. So, what I did is in order for Bob to ALWAYS be able to add numbers to the be the same is Alice, his total of his selection must amount to greater than or equal to Alice's. In order for Alice to maximize her total using k numbers, she chooses the k largest numbers in that set. So basically, the maximum value for k arises when the sum of the last k numbers in a set of successive integers from 1 to n+k is less than the sum of the first n integers. We can set up an inequality where n(n+1)/2>((n+k)(n+k+1)-n(n+1))/2 and using our knowledge that k is equal to 2023-n, solve for n and round up using the quadratic formula. By doing this we get n is equal to 1431 and therefore k is 592 since k = 2023-n. With this you can generalize for any set of successive integers. Edit: Just saw you covered this solution after skipping straight to the answer and that I did this for no reason :(
@@madghostek3026 well in a list of integers, if you have a set of consecutive integers which sums to be greater than another set of consecutive integers, you can subtract enough numbers needed from the larger set until you get equality with the smaller set. The maximum sum of any chosen integers in the first 1431 consecutive integers ranges from 1 to 1431(1432)/2. This is also the range of numbers we can subtract from the first 1431 numbers to get the next 592. Idk exactly how to rigorously prove this as I never do competition math but this was my loose justification.
@@instinx9154 Yeah this is basically the thing everybody struggled with, it's easy to show an upper bound for k by asking "where to split 1,2,...n so that lower half sums up to something more than higher half". But now we don't really know if you can actually pick out numbers in a way to make the sums equal, which is the problem they are asking about. Notice that Alice doesn't need to pick only the largest numbers, let's say she picks 2023, 2022, 2021 and 1. Can you always manage your way without the 1?
@@madghostek3026 oh I gotcha, considering all those edge-cases is what makes it difficult. I just solved it assuming I didn’t have to prove anything lol.
To see how 51 is the answer for the problem: We want a k such that we are ALWAYS sure that any collection of k numbers from {1,2,3,..,100} we can fnd 2 numbers in it such that their sum is odd. Let's look at an even simpler example: We have instead the set {1,2,3,4,5}, what is the smallest k such that any subset of k elements we can find 2 numbers in them with an odd sum? Well k=5 is a candidate since the only 5 element subset of this is just itself so it works, but can k be smaller and still work? Hmm, what if k = 2? No! Why? Because in the case of k = 2 here there is the subset {2,4} such that there is no odd sum in this one :( so k can't be this low... Did you see why {2,4} was important in the argument? Because is it a subset where no odd sum exist and it is made out of only even numbers (duh! No odd sum can exist if we only have even numbers). But it also doesn't work for k = 3 either! Here we have a subset {1,3,5} that has no odd sum (1+3 = 4 , 1+5 =6, 3+5=8 none of which are odd. Well duh! A set with only odd numbers cant have an odd sum!) so k has to be bigger than 3. But does it work for k = 4? Yes, because (as you have seen so far the biggest worst cases possible are the subsets of either all even numbers or all odd numbers) so any subset of 4 elements here must contain an even and odd element since the the biggest worst case is {1,3,5} and having a 4th element adds a even here so all 4 element subsets have at least 1 even and 1 odd number and thus k=4 is the answer here. Back to the original 100 numbers question: Any k below 51 there would always the case of the set { 1, 3, 5, 7, 9, ..., 2k - 1} which is made of only odd numbers and this is always problematic. Since this worst case scenario set can be avoidee be giving it an even number that means there are no bad scenarios for sets with 50 (+1 even) numbers in them. So k = 51 If you try to generalize this you can see that, if A = {1,2,3,4,5,6,...,N} then the approapiate k = floor((N+1)/2)
If we can find a set that sums to alpha' then the complement set (those numbers not in A and not used in the alpha' sum) will sum to alpha as required - so finding a set that sums to alpha' also lets us find a disjoint set that sums to alpha. The value of replacing alpha with alpha' is, if alpha' is smaller than alpha, then it is easier to prove for as we reduce the maximum number of pairs to a low enough level
@@GeorgeFoot That answers the why question, but I'm still not sure about “what allows us to” question. It's kind of hard for me to formulate my question and English is my second language, so sorry in advance. It seems artificial to me to set alpha' to equal 2024a+b. While I totally understand the reasoning behind the formula for alpha, I don't think we can just use the same reasoning here, since alpha' is not necessarily used as a set, whose sum we are interested in (because it can also be equal to S−2alpha).
@@nin0f Yes indeed. alpha' is not a sum of a set yet, just a target value. But if we can find a subset B' of S-A, with B' summing to alpha'=sum(S)-2alpha, then we can let Bob's set B equal S-A-B' and it sums to alpha as required - so using alpha' as a target sum works just as well as alpha, for Bob finding a solution - and whichever or alpha and alpha' is smaller, is easier to provably achieve directly. What he is doing here is really going back and saying, rather than just picking a and b so that 2024a+b=alpha, let's - if alpha'
@@GeorgeFoot I've suspected it was something similar to that, but my brain was already fried at that point of the video to connect the dots myself. So thank you, it all makes sense now!
I actually don't use manim, I just use the Computer Modern font (which is standard). I use a mix of davinci resolve (in the fusion page) and after effects.
@@Polyamathematics oh, that's cool! you're able to get some very beautiful results with that. I associate with manim-produced videos even more because of the music :) great work!
Is it just me or is this the wrong answer. If k=1 and Alice colours just the number one then Bob can’t solve it. So the largest k is should be 0. This is a weird edge case but I can’t see why this shouldn’t be a solution. However I really enjoyed the video and the problem
k=1011 alice colours 1-1011 red (1011) bob colours 1012-2022 blue (1011) it says some so there can be uncoloured numbers so 2023 remains as the only uncoloured one 1011=1011 proof: if k>1011 then alice colours 1-1012 red (1012) bob colours 1013-2023 blue (1011) 1012>1011
4:51 Here's my proof: There are only 50 even and 50 odd numbers in the set, so there are an even number and an odd number in the subset, and their sum is odd For me (an autist), it's easier, although it uses a fact that some people may not know (the oddity of a sum)
@@quantumgaming9180 There are only 50 even numbers. Therefore, the subset must have an odd number. Similarly, it must have an even number. The sum of an even number and an odd number is odd
Edit: failed at reading comprehension, see replies Counter example: Alice chooses the first 1428 integers, they will sum up to 1020306. then if Bob chooses the remaining integers they add up to 1026970. However Bob can remove 1664, 1665, 1667, and 1668 from his choices, making his sum 1020306 - same as Alice's. So k=592 is wrong
The problem is to find the largest interger k such that for *any* choice made by Alice of k numbers, Bob can choose some other numbers with the same sum. What your counter example misses is that, although you provided *one* example where Alice chooses 1428 numbers and Bob is able to choose numbers accordingly, it's not true that Bob could do that for *all* choices of 1428 numbers that Alice could have made (easy to see that if Alice had chosen the last 1428 numbers that would be impossible).
@@samuelleite7488 Thank you, that's what I've been missing. I've re-read the problem statement many times and I still wouldn't read it as "for any choice of k integers", guess I'm bad at English
@@czarelius no problem, statements can be tricky. In this one, I guess the key to understand it is the word " *whenever* Alice colours exactly k numbers...". That would mean you have to account for any possible choice of k numbers.
Wouldnt you just sum all of the numbers and divide by 2 to find the maximum sum? 0.5(2023)^2 +0.5(2023) = 2047276 Work backwards to find how many numbers are needed to surpass 1023638 -----> bob needs 1431/2023 numbers, so K is 592. It is OBVIOUS there will always exist a number you can leave out to obtain the required sum. I think that is what students wrote but the graders had a stick up their ass
By this logic if 2023 was replaced by say 5 Then we're colouring numbers in the set {1,2,3,4,5}. This sums to 15 so 4+5 > 15/2 so here k=1. Not only is it not "obvious" that Bob has a strategy, but it's straight up wrong
@6:27 concerning the expression k² - 4027k + 2023×1012 ; is 4027 a typo? i expect 4047 = (2n+1) for n=2023 using the notation, for any a,b, n el Nat : [ a ; b ] := { x el Nat | a
![Why is this "Fundamental" to Arithmetic? #SoMEpi [5th]](http://i.ytimg.com/vi/46E0-XJuAXs/mqdefault.jpg)
I’ve no problems in math in any placements: schools, universities, but I always been struggled at the olympiads. This questions wants some impossible thinking during short time, which requires that you knows these types of questions and know how they are solving, otherwise you’re just a genius, who can solve any sort of problems in few hours
That’s because they’re used to test the best young mathematicians in the world these problems aren’t meant to be easy and anyone who says they are is probably lying
@@ThoseInterestingStories i mean like the only hard problems are millennium problems and most of the time they do not even have a right answer let alone a proper solving technique known to current principles of mathematics
There are trainings for this type of problems
Math olympiads make no sense, they are the exact opposite of what math should be about
Fun fact:
The person who solved the problem entirely (got 10/10) is Romanian and he also got a max score in the IMO that year. Being Romanian too, I see this guy as the Romanian Jesus of Mathematics, you can show him any problem and he will solve it for.
Anyway, completely coincidentally, the author of the problem is also Romanian, and the original text used "n natural number" instead of "2023", but they changed it before the contest as it probably would have been an overkill.
I'm Romanian too and I completely agree lol
Currently at 7:08 and there is one more constraint on k. If k = 1, Alice can choose 1 or 2 for that one number, and Bob’s task is impossible. So 1 < k < 593
5:00 I solved this a different way: if there are more than 50 numbers in the subset, we know that they can't all be odd / can't all be even (since there are only 50 odds and 50 evens in the original set) - thus, there is at least one odd number and one even number, which we can add to get an odd number.
I feel like this is the natural way of doing it (I know I did). Also interestingly this is actually the same method. However, the lessons learned from the smaller problem stated in the video are from turning your proof into one of pairs and pigeonhole. But I wouldn’t have got the same “tooling” from your (and my own) explanation. Interesting how different interpretations lead to different progressions.
@@williamchurcher9645 for sure!
My thoughts after hearing the problem:
The logical limit would be bΔ = 2023Δ/2, where b = 2023 - k, giving k = 592 (rounded down)
If Alice takes the 592 largest numbers, Bob can take the rest (excluding numbers summing to 1916, to make it equal). If Alice takes the 593 largest numbers, the sum is greater than the sum of the remaining numbers, so Bob cannot make an equal sum.
Intuition says that there will always be plenty of small numbers for Bob to choose from to make the numbers equal, but I'll try to come up with a logical method:
Once Alice chooses her numbers, Bob chooses the remaining numbers from the largest down, until his sum exceeds Alice's sum. He then chooses one extra number, that is the negative of his excess (for example, if his chosen numbers go down to 61, and his sum is now 25 greater than Alice's sum, he adds the number -25 to his set, to even them out). He then reduces the smallest non-negative number to the next available number (e.g. 61 => 60), and then increases the negative number by the difference to keep it even. He then sends the new gap (now at 61 in the example) bubbling up his set, so he can increase the negative number into the positives, until it eventually meets the rest of his set (e.g. [-24, 60, 62, 63] => [-23, 60, 61, 63] => [-22, 60, 61, 62] => [-21, 59, 61, 62] etc).
In the case where Alice chooses the largest 592 numbers, there are ~60 legal spots for the floating number to end in. To prevent this method from working, Alice would have to have to occupy at least 60 of the first ~120 numbers (blocking spots where either the floating number or the bubble would want to be in each iteration). However, doing this would increase the number of Bob's unused numbers faster than she could block them (e.g. if she chooses the largest 591 numbers and then also 120, it would reduce the sum by ~1300, meaning Bob would only have to choose numbers down to ~80, so Alice would now have to block 80 of the first ~160 numbers).
Yes
Legend of the Problem 4
Nice!
challenge: take a shot every time Mr Polyamath says the word greedy
,loi.hjjjjjjjjjjjjjjjjjjjjjjjko'p/;lko
Hi, I’m that guy who made a bunch of problems in the math problem channel in your discord server. Good vid
Loving the videos!
Polyamath posted, hence I stop everything I am doing
I'll figure it out one day
i love your videos
This is a great problem and some interesting solutions, but you jump from one idea to another so quickly and oftentimes briefly mention relevant information or use so much previously defined variables without reminding the viewer what they were, that closer to the end of the video it became unwatchable.
0:15 excuse me sir, Hungary is not a Balkan country (on paper)
The way the problem is presented confused me at first since it didn’t say any k of the set just exactly k. Maybe it’s a standard that these problems are worst case scenarios but I assumed it was alice working to maximize k, in which case she would choose the smaller numbers over the larger.
Generally speaking, olympiad problems give you the absolute least amount of information to be 100% certain what the question is. If the question says "find the greatest integer k such that..." Then they do just mean the result holds at exactly k.
If they wanted it to hold for all values up to k they'd say "find the greatest k such that for all m ≤ k" the result holds.
Awesome video! Here is a nice question for the meticulous viewers: where does the proof fail for k=593?
[Spoiler]
16:20
Fire editing skills ✨ 🔥
Not sure what it is yet, but can't be more than 592. Otherwise Alice would pick the highest numbers and Bob would not be able to get a sum of equal size.
Lol, I got the 1 mark!
Nuts that you already got a sponsor lol
God these are good
20:47 where the hell does that implication come from?!
To my own surprise I actually solved this one. So, what I did is in order for Bob to ALWAYS be able to add numbers to the be the same is Alice, his total of his selection must amount to greater than or equal to Alice's. In order for Alice to maximize her total using k numbers, she chooses the k largest numbers in that set. So basically, the maximum value for k arises when the sum of the last k numbers in a set of successive integers from 1 to n+k is less than the sum of the first n integers. We can set up an inequality where n(n+1)/2>((n+k)(n+k+1)-n(n+1))/2 and using our knowledge that k is equal to 2023-n, solve for n and round up using the quadratic formula. By doing this we get n is equal to 1431 and therefore k is 592 since k = 2023-n. With this you can generalize for any set of successive integers.
Edit: Just saw you covered this solution after skipping straight to the answer and that I did this for no reason :(
This doesn't prove yet that an actual sum exists, you only know that there doesn't exist counter example that sums up to impossible value
@@madghostek3026 well in a list of integers, if you have a set of consecutive integers which sums to be greater than another set of consecutive integers, you can subtract enough numbers needed from the larger set until you get equality with the smaller set. The maximum sum of any chosen integers in the first 1431 consecutive integers ranges from 1 to 1431(1432)/2. This is also the range of numbers we can subtract from the first 1431 numbers to get the next 592. Idk exactly how to rigorously prove this as I never do competition math but this was my loose justification.
@@instinx9154 Yeah this is basically the thing everybody struggled with, it's easy to show an upper bound for k by asking "where to split 1,2,...n so that lower half sums up to something more than higher half". But now we don't really know if you can actually pick out numbers in a way to make the sums equal, which is the problem they are asking about. Notice that Alice doesn't need to pick only the largest numbers, let's say she picks 2023, 2022, 2021 and 1. Can you always manage your way without the 1?
@@madghostek3026 oh I gotcha, considering all those edge-cases is what makes it difficult. I just solved it assuming I didn’t have to prove anything lol.
i love this video
At 20:23 you say we have 2 edge cases with bi=1 or bi=2, but doesn't bi/2-2 imply another 2 edge cases in bi=3 and bi=4 ?
uhhhh i don't get it, why we're taking pairs of numbers and why k=51... i feel bad :C
To see how 51 is the answer for the problem:
We want a k such that we are ALWAYS sure that any collection of k numbers from {1,2,3,..,100} we can fnd 2 numbers in it such that their sum is odd.
Let's look at an even simpler example:
We have instead the set {1,2,3,4,5}, what is the smallest k such that any subset of k elements we can find 2 numbers in them with an odd sum? Well k=5 is a candidate since the only 5 element subset of this is just itself so it works, but can k be smaller and still work? Hmm, what if k = 2? No! Why? Because in the case of k = 2 here there is the subset {2,4} such that there is no odd sum in this one :( so k can't be this low... Did you see why {2,4} was important in the argument? Because is it a subset where no odd sum exist and it is made out of only even numbers (duh! No odd sum can exist if we only have even numbers). But it also doesn't work for k = 3 either! Here we have a subset {1,3,5} that has no odd sum (1+3 = 4 , 1+5 =6, 3+5=8 none of which are odd. Well duh! A set with only odd numbers cant have an odd sum!) so k has to be bigger than 3. But does it work for k = 4? Yes, because (as you have seen so far the biggest worst cases possible are the subsets of either all even numbers or all odd numbers) so any subset of 4 elements here must contain an even and odd element since the the biggest worst case is {1,3,5} and having a 4th element adds a even here so all 4 element subsets have at least 1 even and 1 odd number and thus k=4 is the answer here.
Back to the original 100 numbers question:
Any k below 51 there would always the case of the set { 1, 3, 5, 7, 9, ..., 2k - 1} which is made of only odd numbers and this is always problematic. Since this worst case scenario set can be avoidee be giving it an even number that means there are no bad scenarios for sets with 50 (+1 even) numbers in them. So k = 51
If you try to generalize this you can see that, if A = {1,2,3,4,5,6,...,N} then the approapiate k = floor((N+1)/2)
14:11 what allows us to (or why do we) replace alpha with alpha-dash?
If we can find a set that sums to alpha' then the complement set (those numbers not in A and not used in the alpha' sum) will sum to alpha as required - so finding a set that sums to alpha' also lets us find a disjoint set that sums to alpha.
The value of replacing alpha with alpha' is, if alpha' is smaller than alpha, then it is easier to prove for as we reduce the maximum number of pairs to a low enough level
@@GeorgeFoot That answers the why question, but I'm still not sure about “what allows us to” question. It's kind of hard for me to formulate my question and English is my second language, so sorry in advance.
It seems artificial to me to set alpha' to equal 2024a+b. While I totally understand the reasoning behind the formula for alpha, I don't think we can just use the same reasoning here, since alpha' is not necessarily used as a set, whose sum we are interested in (because it can also be equal to S−2alpha).
@@nin0f Yes indeed. alpha' is not a sum of a set yet, just a target value. But if we can find a subset B' of S-A, with B' summing to alpha'=sum(S)-2alpha, then we can let Bob's set B equal S-A-B' and it sums to alpha as required - so using alpha' as a target sum works just as well as alpha, for Bob finding a solution - and whichever or alpha and alpha' is smaller, is easier to provably achieve directly. What he is doing here is really going back and saying, rather than just picking a and b so that 2024a+b=alpha, let's - if alpha'
@@GeorgeFoot I've suspected it was something similar to that, but my brain was already fried at that point of the video to connect the dots myself. So thank you, it all makes sense now!
“you got a solution in a way we didn’t want so you lose points” is such bs 😭
they didn't lose points because of the way they solved it, the solution just didn't have enough details. see 17:23
a pair of sum b, and a pairs of sum 2024.
What do you use to make your videos, its very similar to veritasiums style
I would say it's very similar to 3blue1brown's style and that polyamath probably uses manim, 3b1b's video making python library
I actually don't use manim, I just use the Computer Modern font (which is standard). I use a mix of davinci resolve (in the fusion page) and after effects.
@@Polyamathematics oh, that's cool! you're able to get some very beautiful results with that. I associate with manim-produced videos even more because of the music :) great work!
Yes some of the music is used by too 3blue1brown!
Is it just me or is this the wrong answer. If k=1 and Alice colours just the number one then Bob can’t solve it. So the largest k is should be 0. This is a weird edge case but I can’t see why this shouldn’t be a solution. However I really enjoyed the video and the problem
The result isn't true for all integers up to 592, it's only true specifically at k = 592.
@@TheArizusok I see. Thank you!
k=1011
alice colours 1-1011 red (1011)
bob colours 1012-2022 blue (1011)
it says some so there can be uncoloured numbers so 2023 remains as the only uncoloured one
1011=1011
proof: if k>1011 then
alice colours 1-1012 red (1012)
bob colours 1013-2023 blue (1011)
1012>1011
it says the sum of the coloured numbers must be equal, not the number of coloured numbers
4:51 Here's my proof: There are only 50 even and 50 odd numbers in the set, so there are an even number and an odd number in the subset, and their sum is odd
For me (an autist), it's easier, although it uses a fact that some people may not know (the oddity of a sum)
I dont seem to understand your proof, can you elaborate more?
@@quantumgaming9180 There are only 50 even numbers. Therefore, the subset must have an odd number. Similarly, it must have an even number. The sum of an even number and an odd number is odd
Edit: failed at reading comprehension, see replies
Counter example: Alice chooses the first 1428 integers, they will sum up to 1020306. then if Bob chooses the remaining integers they add up to 1026970. However Bob can remove 1664, 1665, 1667, and 1668 from his choices, making his sum 1020306 - same as Alice's. So k=592 is wrong
The problem is to find the largest interger k such that for *any* choice made by Alice of k numbers, Bob can choose some other numbers with the same sum.
What your counter example misses is that, although you provided *one* example where Alice chooses 1428 numbers and Bob is able to choose numbers accordingly, it's not true that Bob could do that for *all* choices of 1428 numbers that Alice could have made (easy to see that if Alice had chosen the last 1428 numbers that would be impossible).
@@samuelleite7488 Thank you, that's what I've been missing. I've re-read the problem statement many times and I still wouldn't read it as "for any choice of k integers", guess I'm bad at English
@@czarelius no problem, statements can be tricky. In this one, I guess the key to understand it is the word " *whenever* Alice colours exactly k numbers...". That would mean you have to account for any possible choice of k numbers.
Wouldnt you just sum all of the numbers and divide by 2 to find the maximum sum?
0.5(2023)^2 +0.5(2023) = 2047276
Work backwards to find how many numbers are needed to surpass 1023638 -----> bob needs 1431/2023 numbers, so K is 592. It is OBVIOUS there will always exist a number you can leave out to obtain the required sum. I think that is what students wrote but the graders had a stick up their ass
By this logic if 2023 was replaced by say 5
Then we're colouring numbers in the set {1,2,3,4,5}. This sums to 15 so 4+5 > 15/2 so here k=1.
Not only is it not "obvious" that Bob has a strategy, but it's straight up wrong
@6:27 concerning the expression k² - 4027k + 2023×1012 ; is 4027 a typo? i expect 4047 = (2n+1) for n=2023
using the notation, for any a,b, n el Nat :
[ a ; b ] := { x el Nat | a