here we have problem one from the 2024 IMO International math Olympiad like all the IMO problems it's a really nice problem um being problem one it's more accessible than the other problems is going to take a bit of time and a bit of concentration determination to stick with it and follow the problem through should also say before we get into it Well Done USA for coming first place this year second place uh China third place Korea fourth place India fantastic result and fifth place bellarus all Al huge improvement from that team okay so the problem is to determine all real numbers Alpha such that for every positive integer n floor of alpha plus floor of 2 Alpha plus floor of 3 Alpha Etc all the way up to n Alpha is a multiple of N and this last sence explains that floor function for us so these brackets um indicating the greatest integer less than or equal to for example if we're at Pi uh we round further down to-4 floor of two we just stay at two we floor of 2.9 we round back down to two so almost always the best way to get started with something like this is just to sub in some numbers and see what happens so let's start subbing in some numbers for Alpha we're trying to solve for a real number Alpha um but let's just say if Alpha is a whole number to start with so if Alpha is equal to 1 then you know this thing has to be a multiple of n for all integers n so let's start with n is equal to 1 when Alpha is 1 and N is equal to 1 we just get the floor of alpha which is four of one is just one um one is a multiple of one so we're all good so far um but when we Sub in N is equal to 2 so we don't really need to worry about the floors because floor of a whole number is just um a whole number so floor of one is one um floor of two is two but we get 1 + 2 which is three now three is not a multiple of two so therefore the test fails the sum is not a multiple of n for all n and we can say that Alp equals 1 is definely not a solution so that's good let's try Alpha = 2 now if we Sub in um Alpha = 2 and N = 1 we'll just get 2 when n = 2 we'll get um Alpha + 2 Alpha which is 2 + 4 that is 6 6 is a multiple of 2 keep going um Alpha + 2 Alpha + 3 Alpha is 2 + 4 + 6 that's 12 which is a multiple of 3 keep going again you can see we get that same sum 2 + 4 + 6 well we had 12 from before add a get 20 yes still looking good um add another term so we're adding 5 Alpha which is 10 we get 30 it is a multiple of five so it's looking like um when Alpha is equal to 2 this seems to work that we always get this sum being a multiple of n no matter what the value of n is and we can actually prove that that's always going to be the case when Alpha is equal to 2 in fact when Alpha is equal to any even integer um and we can do so using this formula so the sum of the integers from from 1 to n is always going to be equal to n * n + 1 / 2 now if you haven't seen that formula before I think the easiest way um to explain it to understand it is think about adding the numbers for example from 1 to 5 um what we can do to simplify that is replace all of the numbers with the middle number which is three we haven't changed the sum so we can always do that the average is always going to be uh n + 1 over two the middle number and then we get n * that average number so n * n + 1 / 2 so given that now sum we get Alpha * n * n + 1 / 2 that's our formula if Alpha is even then Alpha over2 is going to be an integer so what do we have we have an integer multiple of n so therefore the sum is always a multiple of n for all even integers n so that's good we've got a set of solutions which is all even integers if Alpha is odd however then this is actually never going to work so if we look at our sum again which is the same Alpha * n n + 1 / 2 if Alpha is odd then it's not divisible by two and if we want this to be a multiple of n then that means then n + 1 would have to be divisible by two but that's not going to work for all n okay so actually if n plus1 is odd then it's not going to be divisible by two we know Alpha is not divisible by two so this thing would not be a multiple of n okay so we're going well so so far we've shown that for integers it always works for all even integers it never works for any odd integer but of course Alpha does not have to be an integer it can be any real number so what happens if Alpha is not an integer again we're going to use our strategy of trying some uh numbers and see what happens so if we try for example um Alpha is a half and then for Nal 1 we just get the floor of a half now the floor of a half we round down to zero actually is a multiple of one okay so we say Z is a multiple of any number because 0 is like 0 times any number if we try Nal 2 so we get floor of alpha plus floor of 2 Alpha we get 0 + 1 which is 1 now one is not a multiple of two so Alpha equal a half is not in our solution set let's try another one say Alpha is equal to 2/3 um when n equal 1 the floor of 2/3 is zero that's fine uh but when Alpha is equal to two we'll get the floor of 2/3 plus the floor of 4/3 0 + 1 again is 1 not a multiple of two um let's try another one like how about something bigger than one like 1 and 3/4 which we could also write as 1.75 or 7 over4 when n = 1 the floor is one that's fine when n = 2 the floor of 2 * Alpha we can work out that the floor is actually three so 1 + 3 is 4 and that actually works when n is equal to 3 so we got Alpha 2 Alpha 3 Alpha well Alpha and 2 Alpha we had that from the previous sum we know that's four okay so all we need to do is work out the floor of 3 Alpha uh 3 * 7 over 4 we can work out that that is just over five um so now we add 4 + 5 to get 9 and actually it still works so you know are we on to something here uh let's try n is equal to 4 from the first three terms we have nine from our previous sum and then we adding on the floor of four Alpha we can work out that that is seven and we get 16 16 is a multiple of four okay but if we try the next one when n is equal to 5 we add on to our 16 the floor of five Alpha and we can work out that we get 16 + 8 which is 24 that is not a multiple of five so our fraction 1 and 3/4 actually fails and it's not a solution so after trying these examples where Alpha is not an integer you might be starting to expect that actually there's not any other Solutions um that are not integers you know besides the even integers we found before um but let's just try one more example so let's take a negative we haven't taken a negative yet so if we try um Alpha is -5 um for n = 1 floor is1 for n = 2 um 2 Alpha is still has a floor of -1 so we still good -2 is a multiple of two um when n is equal to 3 we have our -2 from the first two terms and 3 Alpha is still uh less than well it's still within 1 and 0 so the floor is still -1 um the sum is -3 still works and the same thing is going to keep on happening for n = 4 and N is equal to 5 it's only when we get to n is equal to 6 that that 6 Alpha term becomes now um as larger than one the size is larger than one so the floor of -6 on 5 is actually -2 the whole sum becomes -7 which is not a multiple of six so looking at this example I think really helps because we can see like actually worked fine for the first five integers okay Alpha was um -1 on 5 and for n = 1 2 3 4 5 it worked out fine the sum was a multiple of n but as soon as um n was six this last term became too large so there's actually something going on there which is significant and I think to really nail down what that is it helps to focus on Alpha between say Z and one first so if Alpha is any uh real number between 0 and 1 the floor is always going to be zero now if we look at 2 Alpha okay cuz when Nal 2 we need to do alpha plus um 2 Alpha there's now two possibilities for the floor of 2 Alpha if Alpha is less than half then both Alpha and 2 Alpha are going to have a floor of zero but as soon as you get to a half when we double it um we're going to be one or bigger so if we're bigger than a half the floor is actually going going to be one floor of alpha plus floor of 2 Alpha is going to be 0 + 1 and that is no longer going to be a multiple of two and we can keep on going like this and break um the interval between 0 and 1 into thirs and within that first third you know the flaw are all going to be zero because 3 Alpha would be less than 1 so we still get 0+ 0 plus 0 that's great still a multiple um but as soon as we get Beyond 1/3 well these for these values here um the floor of three Alpha is going to be now 1 okay because 3 * 1/3 would be 1 anything bigger is going to be bigger than one so we'll get 0+ 0 + 1 it's no longer a multiple of three and actually I don't really need to check any of the rest because I know that any Alpha values that are bigger than a half already failed the test for n is equal to two so I've now narrowed down my possible set of solutions to within Z to 1/3 um and you can probably see where this is going now because if I kept on going each time I'm going to be narrowing down that possible interval of solutions for each time n increases by one the width of that interval um of possible solution values is going to be split into two intervals and for the left of those intervals the sum is going to continue to be zero um but for the right of the intervals the sum is going to be 0 plus 0 + 0 + 1 and therefore it's going to fail the width of that interval of possible solutions is going to decrease and it's going to approach zero but if Alpha is a solution it has to be a solution for all values of N and as n approaches Infinity the interval between 0 to one on N is going to approach uh zero there's going to be no Alpha value small enough that it's always less than one on N except zero itself which is an even integer we already know that's a solution um and now we can do a similar thing for the integral between -1 and zero for any Alpha between 1 and0 the floor is always -1 um if we look at 2 Alpha we now split into two intervals um and for the right of those intervals the floor is going to be -1 + -1 that's -2 that's fine um but as soon as we go beyond - one2 2 Alpha then has a floor of2 so -1 - 2 is -3 which fails the test because it's not a multiple of two keep on going like this um and in this time the right of the intervals the one that's closer to zero all it's all negative 1 okay so1 * N is a multiple of n but the left of those intervals is going to be like a sum of 1's and then one -2 on the end well that gives you uh nus1 or negative braet n + 1 which cannot be a multiple of n for all n so now we' actually ruled out any values of alpha between - 1 and positive 1 except for zero and it's looking pretty unlikely that there's any other uh Solutions Alpha besides besides the even integers we found um but we do need to prove it for any of those values that are bigger than one or less than negative for me it helped to continue that number line approach for a few more numbers and what we can notice if we do that the width of possible solutions then is getting narrower around zero and the next one is around two and we already know zero is solution two is a solution four is a solution all even integers so what appears to happen is that the width of the interval around those even integers become smaller and smaller um approaches zero and we're just left with the even integers themselves all right so now it's time to complete the proof we will need a little bit of algebra for this um but hopefully like the intuition and what we've done above is going to be really helpful um just to recap what have we actually shown we've shown that all even integers Alpha are solutions um no odd integers Alpha are solutions and no values of alpha between -1 and 1 besides zero are solutions what we're going to do is basically use what we've done above um for the integer case and like the the fraction case where the fractions were less than one because actually for any real value Alpha we can always write it as the sum of an integer and a fraction less than one uh so using some Al we can write any Alpha as K plus beta where K is an integer and beta is between 0 and 1 um and the good thing is when we multiply by n for n Alpha and take the floor of that again NK is an integer so the the floor becomes n k plus the floor of n beta and we're getting really close now um because if we let the fraction part be negative so let's say um if beta is between negative 1 and 1 we actually can even do better and write our Alpha as the sum of an even integer and beta okay because any real number is either an even number plus a little bit or an even number minus a little bit so let's say Alpha is equal to um e plus beta where e is an even integer and bet is between -1 and 1 um that's great because you know it's still true that the Floor N Alpha is NE e plus the floor of n beta so if we want the sum of alpha floor of alpha plus floor of 2 Alpha plus floor of 3 Alpha Etc we can separate that into the sum of the even integer parts and the sum of the fraction flaws and we already know from above we already proved that the sum of the integer Parts um is always a multiple of n but the the sum of those floors of fractions which are between ne1 and 1 is not a multiple of n okay unless beta is zero so therefore the entire sum cannot be a multiple of n for all n and that completes the proof let's just demonstrate that with an example with a number so for example if Alpha is 5.9 then we write um Alpha as the next even number which is six minus a little fraction uh 0.1 floor of 6 -.1 + 2 * that so 12 -2 + 3 * that which is 18 - 0.3 um Etc and we can split that up into the sum of the integer parts 6 + 12 + 18 Etc all the way up to 6 n um minus those the floors of those fraction parts so 0.1 plus floor of 0.2 plus floor of 0.3 Etc and we know that for the integer part we know that is a multiple of n we proved that above using the sum of our consecutive integers formula and for the fraction part because 0.1 is between 0 and 1 and we already proved above that no values of alpha between 0 and 1 uh actually give you a sum that's a multiple of n all right so that's it I hope that made sense to you thanks for watching to the end and sticking with it um if you have any questions about any part of that or if I said something wrong you want to correct me please feel free to leave a comment and a thumbs up I appreciate that as well okay