Since the LHS is monotonicly increasing I first solved over the integers to get 481 for x=5 and 561 for x=6, thus any solution must be between these. Next I observed that the LHS won't change if we increase x by less than 1/27, but if we increase it by exactly 1/27 then precisely 3 of the terms will go to the next integer so we get an increase of 3. The same argument applies for the next increase by 1/27 etc, but clearly there's no multiple of 3 that will sum with 481 to give 500, thus no solution.
@Andrew Layton how so? Didn't seem so obvious to me. To me it seems that a multiple of 3 would come out of the fractional part, but I thought the integer part of x might balance that out
@Andrew Layton aah yes good point. In that case I suppose a faster way to get to the solution is to compute f(0) (with f being the LHS of the equation) and noticing it's not congruent to 2 (mod 3)
I like to do these questions without first viewing. I got no solution quite quickly and then spent over an hour looking for my mistake. After checking all my arithmetic many times over the light went on in my head and I thought 'may be no solution'. I skipped ahead to the video's end to reassuringly saw Michael write no solution on the board. I have fallen into this trap many times and I always think I must have made a mistake.
Great video, as always. I'm not sure what I enjoy most: Michael Penn's videos, or the fact that the Internet has a space for the creation/sharing/enjoyment of these kinds of videos (instead of -- or maybe in addition to -- all kinds of more trivial garbage).
Hi! Here's a problem that might interest you a lot: Find all integer solutions of the diophantine equation x!+y!=z!. Greetings from Puerto Rico! I always enjoy and love all your videos and problems.
@@MrConverse Surely there aren't any solutions larger than 1,1,2 since if we assume x,y < z then z! = z * (z-1)!, so z! >= z * (x! + y!)/2. Thus x! + y! = z! only if z
if z > y >= x then z! == 0 (mod y!) y! == 0 (mod x!) let y! = x!p let z! = y!q then 1+p = pq left == 1 (mod p) right == 0 (mod p) which implies p = 1 and q = 2 which implies x! = y! z! = 2y! only solution is: x = 0 or 1 y = 0 or 1 z = 2 [Edit: Added factorial symbols I missed on lines 3 & 4]
A quicker way to solve this is to use Hermite's identity for floor summations, you can get directly the equation 80*[x] + 3*[27*(x-[x])] = 419 which is the equation you got at 10:45.
Coincidentally While working on a problem I found out a general formula by myself yesterday which is helpful in this problem. Summation of [x+k/n] where k goes from 0 to (Pn-1) is P[nx]+n(P(P-1)/2), where [ ] represents greatest integer function, P and n are positive integers. This Question is the case when n=27 and p=3 and the Problem reduces to equation [27x]=419/3 which justifies no solution.
Maybe I missed the great integer function? I would appreciate it if you could point out my mistake(I basically just solved as it was and don't know how I should have used greatest integer function)
@@supertester23 If I had to guess your mistake without seeing your solution, I would guess it's that his formula runs from 0 to Pn-1, and in the problem, the implied summation runs from 1 to Pn-1
@@jadegrace1312 hm I just multiplied things out with the given formula, no other steps, I guess I'm still too much of a noob and should have applied the formula in some special way
The only value of x that is ***close*** to yielding a solution from my calculator work, key word being calculator, is around 5.259 repeating but it most likely has a slight complex part added.
I got the same before realizing it was asking for the floor, then I got 5.2392 - 5.2393. (because it floors, it loses the decimals, it has the problem of not adapting well.)
I don't typically watch UA-cam videos fullscreen, so a quick zoom-in to make sure I'm still following along properly helped. So count me as another vote in favor.
I do not. It makes me dizzy when I'm looking at it closely and then it zooms in. I don't quite get it: if your resolution is so low you can't read the writing, zooming in for a bit and then back out isn't going to help. It's also a pretty annoying zoom that speeds up as it goes, rather than staying constant or speeding up and slowing down.
There is easy way to see there is no sol. Since you have 80 expressions in the sum and 500/ 80 is little more than 6 try 5 for the begining and you see forx=5 you get 481 so you need more 19. Then if you try 5 and 1/27 this will get you 1lees 5 but will add you one 8 since 81/27 is exactly 3 so adding 1/27 give you 3 more to this sum. But we need exactly more 19 which is not divided by 3 so there is no solution
@@drilen9872 You definitely made a mistake, 38/8 is smaller than 5, and with 5 the sum of floors is 481 as yoav indicated above :) 5 + 7/27 (142/27) is indeed the correct answer.
This is an interesting scenario since I assume each of us when confronted with questions like this in school had to find some easy to handle solution like: no solution or an solution that you could write down with only a handful of characters but nothing like an arbitrary irrational number. The same is true here. In reality I believe an "ugly" result would be still quite likely.
i have been puzzling over irrational bases for a few years now, there has to be some way that we can deal with the irrationals... rationally (logically).
I have a nice problem for you. I personally found it in a youtube video but maybe you don't know the solution yet: Take a sphere and place 4 points a b c d randomly on its surface. What is the probability of the shape that has the points a b c d as its corners containing the center of the sphere
Removing the Floor function from all the terms, will make LHS = f(x) = ∑[k=1, 80] (x + k/27) = 80x + ½(80·81)/27 = 80x + 40·3 = 80x + 120 Setting that equal to 500, 80x = 380 x = 19/4 = 4.75 Come to think of it, since removing all the Floor functions only increases the overall sum, this shows that the original equation can't have a solution, x, less than 4.75. Fred
@@gurrrn1102 Uh, no, 320 + 120 = 440. Right? Or joke? I too have a math degree. That doesn't mean we can do simple arithmetic - at least one of my professors disdained those who could. I don't think he considered it real mathematics. Fred
@@ffggddss it was an attempt at a joke, I made a mistake and posted it to youtube. Goes to show what happens (at least to me) when I'm expecting a problem to be solved with a big idea
I didn't "believe" that there is no solution, so I tried the problem by hand. I quickly realised that I can solve this by writing a Python script. I applied the brute force method of incrementing x in steps of 1/27. I got close at x = 141/27, with the sum being 499. Each increment of x increases the answer by 3.
Did it like so: Observation 1: the result will be a step function of x, it will step by +3 at all multiples of 1/27, except at integer values of x, when it will only step by +2. Observation 2: f(x+1)=f(x)+80 Observation 3: f(0)=26×0+27×1+27×2=81 So f(5)=481, f(5 1/27)=484, ..f(5 6/27)=499, f(5 7/27)=502. So no value of x results in 500.
Another solution: If x = 0 then the sum is 26*0 + 27*1 + 27*2 = 81. If x increases with 1/27 the sum increases with 3, unless x becomes a new integer then the increase is only 2. If x increases with 1 the sum obviously increases by 80. So for example there are solutions for 81, 84, 87, ..., 159, 161, 164, 167, ..., 239, 241, 244, 247... but not for 500 or 501.
Let n be the integral part of x. The fractional part of x lies within k/27 and (k+1)/27 Hereby 80n + integral part of ((k+1)/27) + .. ...+ integral part of ((k+80)/27) = 500 or 80n + 27 + 2*27 +3k = 500 or 80n + 3k = 419 = 320 +99 so n = 4 , k.=33 is a solution However k= 33 lies outside the domain of k (
@@Maxim.Borodin yes, but the numbers themselves are outside of our representation of them. You just talked about the infinite set of ratios that dont work, the ratio just needs to be calibrated to the base you choose to math in. The floor function runs off of next lower integer, unless you are at an integer. So, in base 6: 5 fifths is the whole thing; 0.(1)*5=0.(5)=1. And, in base 10, i just said 0.2*5=1. You see how its not actually a thing? It is the distortion of the device we are using to look at them. P.s. our current irrational base system is not rigorous enough, and i suspect that its important, as irrationals outnumber the rationals infinitly.
Knowing this 80n+3m =419 I know m is between 0 and 26 I want to get red of n Closest modual to 26 is 20 80n+3m = 419 (mod 20) 3m = 19 (mod 20) 3m= 39 (mod 20) m=13 (mod 20) The only number between 0 and 26 and equal to 13 (mod 20) is 13 m=13 80n + 39 =419 80 n = 380 n= 4.75 but n should be integers so we have no solution
A very general question that I hope you (Michael) address at least partially, or maybe a special case: Show for all N>1, the product from n=1 to N of (3+1/(a_n)), where each a_n is residue class +1 or -1 modulo 6 and a_n > 1, is never equal to a power of two. Tough, but if there's one guy I know who can do it, it's you!! Anyone else feel free to take a shot... I've been stuck for a while
How to amend slightly this equation so that it has a solution? If slightly, then instead od 500 let's put 499... The number should be equal to 1 mod 3 if we aim at floor of x being 5. If we go down to the floor of x being 4 then the sum should be 2 mod 3. For example 428...
I solved this by just figuring x should be somewhat close to the average of the terms(500/80 = 6.25) minus the average of the fractional part(40/27), so I started with x=5 which gives 481. Then noting that increasing x by < 1/27 doesn't change the answer, but 1/27 changes it by 3 so the right answer must be 5 + ((500 - 481) / 3)/27, but since 19/3 is a fraction itself the answer can't exist because the right answer would lie between 5+6/27 and 5+7/27, but because of the floor all of those intermediary values are just the same as 5+6/27.
This reminds me about problems like sin(x)=2. No solution in highschool, but after that you simply use complex numbers as an interpretation and just calculate the solution anyway ;-) But what kind of broader interpretation we could use here? Hmmm.....
Michael, I believe x=19/4 is a solution just using sum of first n natural numbers is n(n+1)/2 and some straightforward algebra. Let me know if I am mistaken. Thx
If you are saying n=4.75, it makes sense (and it should be rejected cuz n is integer). But how do u get x=4.75? Did u mix both up? Remember the floor function works very differently from the normal brackets. You probably thought u could do it like normal brackets. But u can't
@@proexcel123 Why not just count the total sum of the original function, the brackets shouldn't make a difference in a basic sum: 80x +(1+2+..80)/27 = 500 -> x =4,75. Why do I need the floor function in the first place?
@@JohQx3 It's not whether u need it or not. But the original question has all those floor functions stuck with it alr. So u do have to consider those whole solving.
@@JohQx3 The question says ⌊ x + 1/27⌋ not [ x + 1/27] . The "brackets" are for the Floor function. The Floor function is an L and a backwards L. There are no bars at the top of the L. Floor is similar to rounding, but instead of rounding up and rounding down, Floor chops off the entire fractional or decimal part. It's easier to see with decimals. Also, Floor(a + b) is not always equal to Floor(a) + Floor(b)
Overcompliated imho, I hope there is no error in my thought process. To simplify it must be a series of intergers N,N,..,N,N+1,...,N+1,N+2,...,N+3,..., N+3 where N+1 and N+2 occurs 27-times. So its up to head and tail. Total must be 500 as a sum of 80 integers. So its about 6 on average. Its not hard to find out than N must be 5 so total will be 500 (or at least around). So series must look like 55...5666...666777...77788...8. Number of 5 and 8 in the series as a sum is fixed (26 occurenes of 5 and 8), because 6 and 7 are fixed to 27 occurences. If you take 20 fives (and 6 eights) for example the total sum will be 499. If you take one five less and add one eight total sum will be increased by 3. In other words there total can't be 500, but can be 496,499,502,505 etc.
I'm not good with notation but this is the basic structure of the answer I found. 4.75 appears to be a valid solution for X. I noticed 1+80 is 81. 81/27 is 3. 1+80, 2+79, ... 40+41. 40 copies of 81/27. So 3×40=120. X=(500-120)/80. X=380/ 80. X=4 60/80. X=4 3/4 or 4.75. Was I flawed in how I found that answer?
HOMEWORK : Let f(n) be the largest prime factor of n² + 1. Compute the least positive integer n such that f(f(n)) = n. SOURCE : Guts Round of HMMT Spring 2021
SOLUTION *89* Suppose f(f(n)) = n, and let m = f(n). Note that we have mn | (m² + n² + 1). First we find all pairs of positive integers that satisfy this condition, using Vieta root jumping. Suppose m² + n² + 1 = kmn, for some positive integer k. Considering this as a quadratic in m, let the other root (besides m) be c. We have c + m = kn, so c is an integer. Also, mc = n² + 1. So if m > n then c ≤ n. So if we have a solution (m,n) we can find a smaller solution (n,c). In particular, it suffices to find all small solutions to describe all solutions. A minimal solution must have m = n, which gives only m = n = 1. We have that k = 3. Now the recurrence a_0 = a_1 = 1, a_n + a_(n+2) = 3a_(n+1) describes all solutions with consecutive terms. In fact this recurrence gives precisely other Fibonacci number: 1, 1, 2, 5, 13, 34, 89, 233... Checking these terms gives an answer of 89.
Using logic instead of math (though it works out the same): The 1/27...80/27 portion itself becomes 26x0, 27x, 27x2 = 81 The X's integer portion becomes 80x Every additional 1/27th to X will add three more as 26/27th bits become 1. 500-81=419. This fits 5x, remainder 19. 5+6/27 gets you to 499, and it will stay exactly there until 5+7/27 which is 502. So there are zero results.
From the title page, there is no requirement that x is an integer. Easy way to solve is to realize that the sum of the fractional parts of the first and last terms equals 3. Repeat 40 times since there are 80 terms, so 40 * 3 = 120. Now the problem is easy => 80x = 380 => x = 19/4. No solution if x is real.
I noted the minimum x as n+y, where n in the integer part and m is the under the floor part, y € [0,1) i did somehow simpler, starting to find n, the integer part of the real x. so i started similarly, but with with a calculus of m=0. so i had 26 of n, 27 of (n+1) and 27 of (n+2). that meant 80*n+81 to aproximate to 500. n=5 gives 481, n=6 gives 541. so our number n is 5 and x is x=5+y. so the main formula can be rewritten as "80*5+81+{sum of exceeding floors when (y+i/27)>27}=500", which results in "{sum of exceeding floors when (y+i/27)>27}=19" (i is indices of the element, i is from 1 to 80) I will rename the {sum....} as 'S', so 'S' must be 19. It is clear that we search the 'y' part in an interval of [m/27,(m+1)/27), m natural < 27, as Michael demonstrates, but every time we increase 'm' by one the sum 'S' will increase 3 times because 3 members of the sum of the floors will shift from 0 to 1. Why? because 80 (number of terms) is 81-1, and 81=3*27. So, when 'y' gets incremental with 1/27 we will see 'S' adding two more 3. Ok, 80 means three sets, of 26 terms separated by two integers, but for m between 1 and 26 the sum 'S' will increase with the step of 3. This results that 'S' is divisible by 3, but it doesn't be the 19, as is requested. we can have S=18 for m=6, wich makes x=5+6/27 and the original sum as 499, and if we have x=5+7/27 then the original sum becomes 502, an increase with 3, as stated above. but there is no real number for which this sum is 500, because we always have 3 sets of 27 components with each one increasing by 1, therefore the sum will increment itself with 3.
To solve this we first need to understand what the constraint there are. 1. x is a Real Number and x is the same for each term. 2. there are 80 terms being added because the numerator increases by 1. 3. Floor function is used. Which basically means the decimal part of each term is discarded. Combining points 1 and 2 solves the x part as 80x Point 3 shows that a period forms from the fraction. So every 27th term will give a whole number. 1-26 Floor function reduces the fraction part of term to zero for each term At 27 it equals one (27/27) 28-53 Floor function reduces the fraction part of term to one for each term At 54 it equals two (54/27) 55-80 Floor function reduces the fraction part of term to two for each term Putting these together: (26 x 0) + 1 + (26 x 1) + 2 + (26 x 2) = 81 Combining these parts fives is 80x + 81 = 500 or 80x = 419 From this you can see that x cannot make a whole number, therefore there are no solitons the constraint in the question.
Let the sum of the first 27 terms by y. Then y is an integer, but y+(y+27)+(y+54)=500, and 3 divides the LHS but doesn't divide the RHS. So there's no such x. QED (Quite Easily Done)
Why not just do it in a much simpler way. The 1/27 up to 26/27 gives x as only useful value. The rest is erased by floor thing. 27/27 up to 53/27 gives 1+x as the rest. 54/27 up to 80/27 gives x+2. So we got. 26x + 27(1+x) + 27(2+x) = 500. -> 80x = 393. Which doesn't have a whole number solution.
If x = n + 1/27 (n is a integer). Than the sum we get is 25n + 27(n+1) +27(n+2) + n + 3 = 500. Because floor(n + 1÷27 + 26/27) = x + 1 -> 80n = 419 For 0
This soooo reminds me why I hated math. (Not quite true.) I was actually very, very good at math ... and would spend all weekend working on a problem (even waking up in the middle of the night to finish a problem with no simple solution). I would be the only student to show up with a solution. Unfortunately I was a Chem major. Still, my math was good enough for me to try to solve a chemistry problem in class that had no solution - and announced this fact as I laughed and explained to the professor that he reversed engineered the problem wrong. (How do you initiate a reaction with a negative mass?)
Check again, these arent normal brackets. Besides if you do this exactly the way you described it you end up with 461 (which makes sense because with x=5 you get 481 so x>5 is a must.) I think the way you calculated this is by treating this like normal brackets, in this case you would get x=4,78...
My bad, did not note it was a floor function prob. Should have known the problem would not be easy if you were taking the time to present it! At least we know answer is x=19/4 without the floor notation
Good one; my approach was slightly different, as this term just begged to be completed :) I add floor(x) to both sides and get: floor(x) + floor(x+1/27) + ... + floor(x+80/27) = 500 + floor(x) Then I group them in three of equal fractional part and get: 3 * (floor(x) + floor(x+1/27) + ... + floor(x+26/27)) + (1 + 2) * 27 = 500 + floor(x) Hm. What is this sum of floors? Obviously it's actually floor(27 x)! 3 * floor(27 x) = 419 + floor(x) Substitute y = 27 x. 3 * floor(y) - floor(y / 27) = 419 "Interesting". Whenever y crosses a boundary of integers, the LHS term increases by 2 or 3. There is no other case when the LHS term increases, and it's also obviously monotonous. This makes it excellent for guessing. Well, y=141 yields 418 < 419 and y=142 yields 421 > 419, so there is no solution. How did I know to guess around y=141? Well, I ignored the floor and solved for y, and got 141.4125.
Hey! I was looking at it and thought about this: Can you sum all X (i assumed that there is 80X because there is 80 terms) and the fractions but doint it the first fraction with the last getting 40 terms of (81/27) having: 80x + (1+80)/27 + (2+79)/27... 80x + (40)(81/27) = 500 80x + (40)(3) = 500 80x = 380 x = 38/8 = 4.75 Or there is a error on my process
This would be correct, if those were regular brackets. But those brackets represent the floor function, and you can't just add what's inside them together like you did. For example, [3.5]+[2.5]=3+2=5, and NOT 3.5+2.5=6.
Can we solve it by Arithmetic progression we know the value of a (first term) and we can find common difference which is 1/27 and we know the number of terms and then we also know the sum of Arithmetic progression Sn=n/2(2a+(n-1) d. Pls answer. Thx
Closest could be x~5.259259... (and it gives 499). But no solution that gives 500. Graphing could show this function rises with step of 3 (some cases with step of 1 or 2) and if step value is not solution, then no solutions. What should this be? Primefinder?
As you, I didn't notice the floor operation (parenthesis and floor sign are so close looking). But for X I get a different solution from yours . Are you sure that X=15.5 is correct ? I get 19/4.
@@cret859 Yes, I made a mistake because I did it in bed a bit sleepy. I did this: There are 80 elements, each one is x + a/27, with a from 1 to 80. You can make 40 pairs that sum 2x+81/27 (you take the 1/27 and the 80/27, up to 40/27 + 41/27). That is 40(2x +3) = 500. Then is simple 2x+3 = 12.5 2x = 15.5 (here I was, but I forgot the 2x) x = 7.75 BUT that is not considering the floor operation xD
suppose x is positive, all the terms are grader than x, so x can't be greater than 500/80. One can reason the same way for x negative (notice that the floor is of -3.5 is not -3 but -4). This imply that the solution is restricted. I don't know if this helps or if I misunderstood your question. Best regards,
1/27+...+80/27 = 81x80/54=120. floor(1/27)+...+floor(80/27)=26x0+27x1+27x2=81. That's 39 difference. 39=0(mod 3). It means 500 - 80.floor(x) = 0(mod3). But x is between 5 and 6 (because 5x80+81=481, 6x80+81=561). So floor(x)=5. So there are no soutions. Took me roughly 1 minute mentally. It's the exact same thing as Binary Agenda's reasoning, in fact, just expressed a bit differently. And all in all it s the same thing Michael Penn does in the video, but in a much more analytical and annoying way :). The 500 - 80.floor(x) = 0 (mod 3) is why Michael found n=floor(x)=4 as a possible solution (but with m too large).
Hi Michael. I' m italian and follow your lessons with interest. Could you solve this problem ? Find all integer solutions of the equation : ax^2+5x+2=0 . Thanks for your courtesy .
The discriminant needs to be gte zero if there are to be any real solutions, so 25 - 8a >= 0. So checking on the first few values, we get (a=3, x=-1) ; (a=2, x=-2) ; (a=-3, x=2) ; (a=-7, x=1). And these are apparently all, at least when i interpret the question as saying that a is an integer as well, not just x. If that's not the case, we get infinite many solutions, including: (a=13/9, x=-3) ; (a=9/8, x=-4) ; (a=23/25, x=-5) ; (a=7/9, x=-6) ; etc.
@@irrelevant_noob thanks . This Is the same result that i obtain with wolfram alfa but i wanted a proceedings or the operation that i must do for have the result . Or a generalization of calculus . Ok thanks
@@fabiopompei317 no calculus needed, just a bit of algebra: to get x=n we need ... [ plugging in the quadratic formula: -5 +/- sqrt ( D ) / 2a = n , so +/- sqrt( D ) = 2an + 5 and squaring ... so: ] a = -(5n + 2) / n^2. Guess that was over-complicating things, could've gotten this just by plugging in x=n into the original ax^2+5x+2. ^^ PS "only over" WHICH integer? o.O
floor(x+(an-b)/n) where we take a,b,n,x as integers >0, the result will be x+a-1. Quickly we can figure that the sum will be 26(x+0)+27(x+1)+27(x+2)=500, simplify, 80x+81=500. We end up with x = 419/80. But 80 ł 419 as a matter of fact 419 is a prime so no integer solutions
No, the floor value of x is equal to the greatest integer lower than or equal to x. Even if x is very very close to 4, its floor value will be 3. And for negative numbers, that's the same thing ; so the floor value of -3.2 for example is -4, 'cause -4 ≤ -3.2
For those who thought like me "can't we write it as 80x +(81*40)/27=500 ?". the answer is -What is floor functions? not that i know what it is either. :(
1. The thumbnail shows groupings of two elements, NOT groups of floors! So: 80x+(80⋅81)/(2⋅27)=500 → 80x+120=500 → 80x=380 → x=4.75 Note: I have NEVER seen floor written so sides are connected underneath. 2. Say they are groups of floors. So: 26(x+0)+27(x+1)+27(x+2)=500 → 80x+27+54=500 → 80x=419 → x=5.2375 Oh... I see... JavaScript: sum=0; x=419/80; for(i=1;i
@@gurrrn1102 If this is directed at me - probably best not to speak when you have nothing intelligent to add. If this is directed at the channel owner - you are correct!
Since the LHS is monotonicly increasing I first solved over the integers to get 481 for x=5 and 561 for x=6, thus any solution must be between these. Next I observed that the LHS won't change if we increase x by less than 1/27, but if we increase it by exactly 1/27 then precisely 3 of the terms will go to the next integer so we get an increase of 3. The same argument applies for the next increase by 1/27 etc, but clearly there's no multiple of 3 that will sum with 481 to give 500, thus no solution.
me too
Wow, that method seems very intuitive after familiarising myself with the problem via Michael’s solution
@Andrew Layton how so? Didn't seem so obvious to me. To me it seems that a multiple of 3 would come out of the fractional part, but I thought the integer part of x might balance that out
@Andrew Layton aah yes good point. In that case I suppose a faster way to get to the solution is to compute f(0) (with f being the LHS of the equation) and noticing it's not congruent to 2 (mod 3)
Yeah that’s how I did it too, although being picky I think you meant monotonically nondecreasing
I like to do these questions without first viewing. I got no solution quite quickly and then spent over an hour looking for my mistake. After checking all my arithmetic many times over the light went on in my head and I thought 'may be no solution'. I skipped ahead to the video's end to reassuringly saw Michael write no solution on the board. I have fallen into this trap many times and I always think I must have made a mistake.
Great video, as always. I'm not sure what I enjoy most: Michael Penn's videos, or the fact that the Internet has a space for the creation/sharing/enjoyment of these kinds of videos (instead of -- or maybe in addition to -- all kinds of more trivial garbage).
I was doing this in my head… but I didn’t realize there were floors on the numbers. I was looking at the video picture.
Hi! Here's a problem that might interest you a lot:
Find all integer solutions of the diophantine equation x!+y!=z!.
Greetings from Puerto Rico! I always enjoy and love all your videos and problems.
Well, there are the solutions x=0 or 1, y=0 or 1, z=2. Are there others?
@@MrConverse Surely there aren't any solutions larger than 1,1,2 since if we assume x,y < z then z! = z * (z-1)!, so z! >= z * (x! + y!)/2. Thus x! + y! = z! only if z
Nice! Otro Puertorro. Althought I suck at math.
if z > y >= x
then
z! == 0 (mod y!)
y! == 0 (mod x!)
let y! = x!p
let z! = y!q
then
1+p = pq
left == 1 (mod p)
right == 0 (mod p)
which implies p = 1
and q = 2
which implies
x! = y!
z! = 2y!
only solution is:
x = 0 or 1
y = 0 or 1
z = 2
[Edit: Added factorial symbols I missed on lines 3 & 4]
Shouldn't a diophantine equation be a polynomial equation? I don't think this is one, but that definition. Not to nitpick, just wondering.
A quicker way to solve this is to use Hermite's identity for floor summations, you can get directly the equation 80*[x] + 3*[27*(x-[x])] = 419 which is the equation you got at 10:45.
Thank you for your videos and for sharing the love and enjoyment of math! Truly appreciated!!!
If someone just want to plug & play with some javascript
function f(x) {
let result = 0;
let iterator = 1;
while (iterator
Coincidentally While working on a problem I found out a general formula by myself yesterday which is helpful in this problem.
Summation of [x+k/n] where k goes from 0 to (Pn-1) is P[nx]+n(P(P-1)/2), where [ ] represents greatest integer function, P and n are positive integers.
This Question is the case when n=27 and p=3 and the Problem reduces to equation [27x]=419/3 which justifies no solution.
Hey I got 81x+81, how should I have used your formula diferently?
Maybe I missed the great integer function? I would appreciate it if you could point out my mistake(I basically just solved as it was and don't know how I should have used greatest integer function)
n=27 here so 3[27x]+3(27*13) is what you get by the P[nx]+n(P(P-1)/2) part, which doesn't look quite right
@@supertester23 If I had to guess your mistake without seeing your solution, I would guess it's that his formula runs from 0 to Pn-1, and in the problem, the implied summation runs from 1 to Pn-1
@@jadegrace1312 hm I just multiplied things out with the given formula, no other steps, I guess I'm still too much of a noob and should have applied the formula in some special way
What a great problem!
Thank you, professor.
(Also, I quite liked the zooms.)
That was fun to watch! I don't deal with the floor function much; the technique of breaking the numberline down into parts was educational for me.
The only value of x that is ***close*** to yielding a solution from my calculator work, key word being calculator, is around 5.259 repeating but it most likely has a slight complex part added.
Any number in the range [114/27, 116/27) will give you 499. Your number is in that range but there’s an infinite number of solutions that give you 499
[141/27, 143/27)
I got the same before realizing it was asking for the floor, then I got 5.2392 - 5.2393.
(because it floors, it loses the decimals, it has the problem of not adapting well.)
Great problem! Great presentation! I don't like the zooming in.
No it's nice actually
13:11 I like the zoom but I’m not sure a lot of people will like it
👋🏽
I am not super worried. I make so many videos that sometimes I have to play around with stuff just for practice.
I don't typically watch UA-cam videos fullscreen, so a quick zoom-in to make sure I'm still following along properly helped. So count me as another vote in favor.
on the other hand, if you change the sum to 502, then you get a single solution of n=5, m=7. Lol
Here you technically have infinite solutions, specifically the interval [5 + 6/27, 5 + 7/27) for x
@@kireetpanuganti2168 well in this case it's the entire interval [5 + 7/27, 5 + 8/27). What you've written gives us the total of 499
Thanks to the viewer who tried to trick you. I wouldn't have seen this beautiful problem otherwise.
the engineering point of view: if it exists, it's probably between 5.259259259258988 and 5.259259259259835.
Lol I like the new zoom
I do not. It makes me dizzy when I'm looking at it closely and then it zooms in.
I don't quite get it: if your resolution is so low you can't read the writing, zooming in for a bit and then back out isn't going to help.
It's also a pretty annoying zoom that speeds up as it goes, rather than staying constant or speeding up and slowing down.
@@ZipplyZane nah
My name is Michael and this popped up on my feed and I thought youtube was talking to me.
There is easy way to see there is no sol. Since you have 80 expressions in the sum and 500/ 80 is little more than 6 try 5 for the begining and you see forx=5 you get 481 so you need more 19. Then if you try 5 and 1/27 this will get you 1lees 5 but will add you one 8 since 81/27 is exactly 3 so adding 1/27 give you 3 more to this sum. But we need exactly more 19 which is not divided by 3 so there is no solution
So now change the question to 502 and the answer is x=142/27
@@yoav613 Actually i got x = 38/8 but idk if i made a mistake :$
@@drilen9872 You definitely made a mistake, 38/8 is smaller than 5, and with 5 the sum of floors is 481 as yoav indicated above :) 5 + 7/27 (142/27) is indeed the correct answer.
@@thetom341 Well no the question is to find all x ∈ ℝ, not just one that works. So 'the answer' would be 142/27
I found answers, now trying to find the flaw in my logic. It was a brilliant proof, but too long to include in this margin.
i see what you did there. just wanted to let you know :D
This is an interesting scenario since I assume each of us when confronted with questions like this in school had to find some easy to handle solution like: no solution or an solution that you could write down with only a handful of characters but nothing like an arbitrary irrational number. The same is true here. In reality I believe an "ugly" result would be still quite likely.
i have been puzzling over irrational bases for a few years now, there has to be some way that we can deal with the irrationals... rationally (logically).
I have a nice problem for you.
I personally found it in a youtube video but maybe you don't know the solution yet:
Take a sphere and place 4 points a b c d randomly on its surface.
What is the probability of the shape that has the points a b c d as its corners containing the center of the sphere
X=4. Easy.
Too bad I didn’t notice the floor signs.
Removing the Floor function from all the terms, will make
LHS = f(x) = ∑[k=1, 80] (x + k/27) = 80x + ½(80·81)/27 = 80x + 40·3 = 80x + 120
Setting that equal to 500,
80x = 380
x = 19/4 = 4.75
Come to think of it, since removing all the Floor functions only increases the overall sum, this shows that the original equation can't have a solution, x, less than 4.75.
Fred
@@ffggddss I got the 120, 320 + 120 = 500, right?
Right?
I have a maths degree 😳
@@gurrrn1102 Uh, no, 320 + 120 = 440. Right? Or joke?
I too have a math degree. That doesn't mean we can do simple arithmetic - at least one of my professors disdained those who could. I don't think he considered it real mathematics.
Fred
@@ffggddss it was an attempt at a joke, I made a mistake and posted it to youtube. Goes to show what happens (at least to me) when I'm expecting a problem to be solved with a big idea
No matter how sarcastic you try to be, somebody is going to take you at face value.
Sobering, isn't it?
Mission failed. We'll get him next time.
get integers out of floor function, you get: 3×sum(floors) + 3×27 = 500
3×integer = 419
contradiction
done
I didn't "believe" that there is no solution, so I tried the problem by hand. I quickly realised that I can solve this by writing a Python script. I applied the brute force method of incrementing x in steps of 1/27. I got close at x = 141/27, with the sum being 499. Each increment of x increases the answer by 3.
I was also dubious that there was no solution so I graphed it!
Did it like so:
Observation 1: the result will be a step function of x, it will step by +3 at all multiples of 1/27, except at integer values of x, when it will only step by +2.
Observation 2: f(x+1)=f(x)+80
Observation 3: f(0)=26×0+27×1+27×2=81
So f(5)=481, f(5 1/27)=484, ..f(5 6/27)=499, f(5 7/27)=502. So no value of x results in 500.
Another solution: If x = 0 then the sum is 26*0 + 27*1 + 27*2 = 81. If x increases with 1/27 the sum increases with 3, unless x becomes a new integer then the increase is only 2. If x increases with 1 the sum obviously increases by 80. So for example there are solutions for 81, 84, 87, ..., 159, 161, 164, 167, ..., 239, 241, 244, 247... but not for 500 or 501.
My career in data-analysis has destroyed my theoretical brain. I graphed it and saw the constant 500 was in a jump in the function.
If it stopped at 79 instead of 80, then x in [5+9/27, 5+10/27) would be all the solutions.
Let n be the integral part of x.
The fractional part of x lies within
k/27 and (k+1)/27
Hereby
80n + integral part of ((k+1)/27)
+ .. ...+ integral part of ((k+80)/27)
= 500
or 80n + 27 + 2*27 +3k = 500
or 80n + 3k = 419 = 320 +99
so n = 4 , k.=33 is a solution
However k= 33 lies outside the domain of k (
It might be a weird question... but if 0.(9) = 9/9 = 1, is floor(0.(9))) 1 or 0? It must be 1, right?
Yep, as said at 0:39 "if you're already at an integer, you don't really need to go down [at all]" ;-)
this is only a problem in base 10. convert the question to base 9 and the logic becomes apparent.
@@MrKahrum incorrect, this thing stands in every base, so for example in base 6 1/5=.(1), in base 9 3/8=.(3) and so on
@@Maxim.Borodin yes, but the numbers themselves are outside of our representation of them. You just talked about the infinite set of ratios that dont work, the ratio just needs to be calibrated to the base you choose to math in. The floor function runs off of next lower integer, unless you are at an integer. So, in base 6: 5 fifths is the whole thing; 0.(1)*5=0.(5)=1. And, in base 10, i just said 0.2*5=1. You see how its not actually a thing? It is the distortion of the device we are using to look at them.
P.s. our current irrational base system is not rigorous enough, and i suspect that its important, as irrationals outnumber the rationals infinitly.
@@MrKahrum yeah, I agree, forgot that floor doesnt work like that, floor(xy)=/=floor(x)floor(y)
Drove me crazy. But I was hoping for no solution since all solutions I had were with commas
Knowing this
80n+3m =419
I know m is between 0 and 26
I want to get red of n
Closest modual to 26 is 20
80n+3m = 419 (mod 20)
3m = 19 (mod 20)
3m= 39 (mod 20)
m=13 (mod 20)
The only number between 0 and 26 and equal to 13 (mod 20) is 13
m=13
80n + 39 =419
80 n = 380
n= 4.75 but n should be integers so we have no solution
Ooh that problem looks really cool
Awesome choice and video
You here!
A very general question that I hope you (Michael) address at least partially, or maybe a special case:
Show for all N>1, the product from n=1 to N of (3+1/(a_n)), where each a_n is residue class +1 or -1 modulo 6 and a_n > 1, is never equal to a power of two. Tough, but if there's one guy I know who can do it, it's you!! Anyone else feel free to take a shot... I've been stuck for a while
How to amend slightly this equation so that it has a solution? If slightly, then instead od 500 let's put 499... The number should be equal to 1 mod 3 if we aim at floor of x being 5.
If we go down to the floor of x being 4 then the sum should be 2 mod 3. For example 428...
First of all 80 floors increasing implies 80x
I mean, it says find all, and all is none. So there is an answer to the problem, and it's none, right?
Ø
I solved this by just figuring x should be somewhat close to the average of the terms(500/80 = 6.25) minus the average of the fractional part(40/27), so I started with x=5 which gives 481. Then noting that increasing x by < 1/27 doesn't change the answer, but 1/27 changes it by 3 so the right answer must be 5 + ((500 - 481) / 3)/27, but since 19/3 is a fraction itself the answer can't exist because the right answer would lie between 5+6/27 and 5+7/27, but because of the floor all of those intermediary values are just the same as 5+6/27.
Very good.
But always disappointed when there is no solution.
I feel the opposite way. I love when a problem has no solutions because it kind of defy our strategies and makes us think again if we did all right.
This reminds me about problems like sin(x)=2. No solution in highschool, but after that you simply use complex numbers as an interpretation and just calculate the solution anyway ;-)
But what kind of broader interpretation we could use here? Hmmm.....
I thought I found a solution by there was a mistake in my computation 😑
Michael, I believe x=19/4 is a solution just using sum of first n natural numbers is n(n+1)/2 and some straightforward algebra. Let me know if I am mistaken. Thx
If you are saying n=4.75, it makes sense (and it should be rejected cuz n is integer). But how do u get x=4.75? Did u mix both up?
Remember the floor function works very differently from the normal brackets. You probably thought u could do it like normal brackets. But u can't
@@proexcel123 Why not just count the total sum of the original function, the brackets shouldn't make a difference in a basic sum: 80x +(1+2+..80)/27 = 500 -> x =4,75. Why do I need the floor function in the first place?
@@JohQx3 It's not whether u need it or not. But the original question has all those floor functions stuck with it alr. So u do have to consider those whole solving.
@@JohQx3 The question says ⌊ x + 1/27⌋ not [ x + 1/27] . The "brackets" are for the Floor function. The Floor function is an L and a backwards L. There are no bars at the top of the L.
Floor is similar to rounding, but instead of rounding up and rounding down, Floor chops off the entire fractional or decimal part. It's easier to see with decimals.
Also, Floor(a + b) is not always equal to Floor(a) + Floor(b)
@@steveodonnell3030 Oh, thanks for explaining :)
Overcompliated imho, I hope there is no error in my thought process. To simplify it must be a series of intergers N,N,..,N,N+1,...,N+1,N+2,...,N+3,..., N+3 where N+1 and N+2 occurs 27-times. So its up to head and tail. Total must be 500 as a sum of 80 integers. So its about 6 on average. Its not hard to find out than N must be 5 so total will be 500 (or at least around). So series must look like 55...5666...666777...77788...8. Number of 5 and 8 in the series as a sum is fixed (26 occurenes of 5 and 8), because 6 and 7 are fixed to 27 occurences. If you take 20 fives (and 6 eights) for example the total sum will be 499. If you take one five less and add one eight total sum will be increased by 3. In other words there total can't be 500, but can be 496,499,502,505 etc.
I love how smart this comment section is.
We can see
419 = 80n+3m and 0
I'm not good with notation but this is the basic structure of the answer I found. 4.75 appears to be a valid solution for X. I noticed 1+80 is 81. 81/27 is 3. 1+80, 2+79, ... 40+41. 40 copies of 81/27. So 3×40=120. X=(500-120)/80. X=380/ 80. X=4 60/80. X=4 3/4 or 4.75. Was I flawed in how I found that answer?
HOMEWORK : Let f(n) be the largest prime factor of n² + 1. Compute the least positive integer n such that f(f(n)) = n.
SOURCE : Guts Round of HMMT Spring 2021
can i get some hint?
I think n must be prime
Solution:
wait until he post
SOLUTION
*89*
Suppose f(f(n)) = n, and let m = f(n).
Note that we have mn | (m² + n² + 1). First we find all pairs of positive integers that satisfy this condition, using Vieta root jumping.
Suppose m² + n² + 1 = kmn, for some positive integer k. Considering this as a quadratic in m, let the other root (besides m) be c. We have c + m = kn, so c is an integer. Also, mc = n² + 1. So if m > n then c ≤ n. So if we have a solution (m,n) we can find a smaller solution (n,c). In particular, it suffices to find all small solutions to describe all solutions. A minimal solution must have m = n, which gives only m = n = 1. We have that k = 3.
Now the recurrence a_0 = a_1 = 1, a_n + a_(n+2) = 3a_(n+1) describes all solutions with consecutive terms. In fact this recurrence gives precisely other Fibonacci number: 1, 1, 2, 5, 13, 34, 89, 233... Checking these terms gives an answer of 89.
@@goodplacetostop2973 zilch
Using logic instead of math (though it works out the same):
The 1/27...80/27 portion itself becomes 26x0, 27x, 27x2 = 81
The X's integer portion becomes 80x
Every additional 1/27th to X will add three more as 26/27th bits become 1.
500-81=419. This fits 5x, remainder 19. 5+6/27 gets you to 499, and it will stay exactly there until 5+7/27 which is 502. So there are zero results.
13 minute math edging video. I was getting all excited towards then end and.... nope, nothing, no solution.
Check my comment added today
From the title page, there is no requirement that x is an integer. Easy way to solve is to realize that the sum of the fractional parts of the first and last terms equals 3. Repeat 40 times since there are 80 terms, so 40 * 3 = 120. Now the problem is easy => 80x = 380 => x = 19/4. No solution if x is real.
Excuse me, if x is an integer. :)
I noted the minimum x as n+y, where n in the integer part and m is the under the floor part, y € [0,1)
i did somehow simpler, starting to find n, the integer part of the real x. so i started similarly, but with with a calculus of m=0.
so i had 26 of n, 27 of (n+1) and 27 of (n+2). that meant 80*n+81 to aproximate to 500. n=5 gives 481, n=6 gives 541. so our number n is 5 and x is x=5+y.
so the main formula can be rewritten as "80*5+81+{sum of exceeding floors when (y+i/27)>27}=500",
which results in "{sum of exceeding floors when (y+i/27)>27}=19"
(i is indices of the element, i is from 1 to 80)
I will rename the {sum....} as 'S', so 'S' must be 19.
It is clear that we search the 'y' part in an interval of [m/27,(m+1)/27), m natural < 27, as Michael demonstrates, but every time we increase 'm' by one the sum 'S' will increase 3 times because 3 members of the sum of the floors will shift from 0 to 1. Why? because 80 (number of terms) is 81-1, and 81=3*27. So, when 'y' gets incremental with 1/27 we will see 'S' adding two more 3. Ok, 80 means three sets, of 26 terms separated by two integers, but for m between 1 and 26 the sum 'S' will increase with the step of 3.
This results that 'S' is divisible by 3, but it doesn't be the 19, as is requested. we can have S=18 for m=6, wich makes x=5+6/27 and the original sum as 499, and if we have x=5+7/27 then the original sum becomes 502, an increase with 3, as stated above. but there is no real number for which this sum is 500, because we always have 3 sets of 27 components with each one increasing by 1, therefore the sum will increment itself with 3.
To solve this we first need to understand what the constraint there are.
1. x is a Real Number and x is the same for each term.
2. there are 80 terms being added because the numerator increases by 1.
3. Floor function is used. Which basically means the decimal part of each term is discarded.
Combining points 1 and 2 solves the x part as 80x
Point 3 shows that a period forms from the fraction. So every 27th term will give a whole number.
1-26 Floor function reduces the fraction part of term to zero for each term
At 27 it equals one (27/27)
28-53 Floor function reduces the fraction part of term to one for each term
At 54 it equals two (54/27)
55-80 Floor function reduces the fraction part of term to two for each term
Putting these together: (26 x 0) + 1 + (26 x 1) + 2 + (26 x 2) = 81
Combining these parts fives is 80x + 81 = 500 or 80x = 419
From this you can see that x cannot make a whole number, therefore there are no solitons the constraint in the question.
In the last part you can use the chinese remainder theorem to find the solutions (if you started with a much larger solution than 500)
Hi, Another way of saying the same thing (but a little bit different). Let’s suppose x has an integer part, I, and a decimal part, d (and so 0
Let the sum of the first 27 terms by y. Then y is an integer, but y+(y+27)+(y+54)=500, and 3 divides the LHS but doesn't divide the RHS. So there's no such x. QED (Quite Easily Done)
Except that there are 80, not 81 terms. But see my solution in my own post for an approach based on your idea.
Why not just do it in a much simpler way.
The 1/27 up to 26/27 gives x as only useful value. The rest is erased by floor thing.
27/27 up to 53/27 gives 1+x as the rest.
54/27 up to 80/27 gives x+2.
So we got. 26x + 27(1+x) + 27(2+x) = 500. -> 80x = 393. Which doesn't have a whole number solution.
If x = n + 1/27 (n is a integer).
Than the sum we get is 25n + 27(n+1) +27(n+2) + n + 3 = 500. Because floor(n + 1÷27 + 26/27) = x + 1
-> 80n = 419
For 0
I did it like this, though I'm not sure if I'm correct
(Edit: I'm wrong)
Let floor(x) = n (integer)
So floor(x + k
floor(x+k, k
@@anastasissfyrides2919 ah I see thanks!
This soooo reminds me why I hated math. (Not quite true.) I was actually very, very good at math ... and would spend all weekend working on a problem (even waking up in the middle of the night to finish a problem with no simple solution). I would be the only student to show up with a solution.
Unfortunately I was a Chem major. Still, my math was good enough for me to try to solve a chemistry problem in class that had no solution - and announced this fact as I laughed and explained to the professor that he reversed engineered the problem wrong. (How do you initiate a reaction with a negative mass?)
There might not be solutions, but this is bounded, right?
Replacing 500 by 499 makes a big difference concerning number of solutions :)
?
@@NoNameAtAll2 Just try to solve it with 499 on the RHS. Hint: you should find infinitely many solutions. :)
floor(x+a/27) has only 3 values for n-1/27
does 5.259259259259259259259...(overbar) work?
I'm wondering what the floor of 3.99 recurring evaluates to
4, since in the limit 3.(9) is just 4, so you basically take the floor of an integer
x = 4,75, you can solve the brackets then add everything up and you get this result
Check again, these arent normal brackets. Besides if you do this exactly the way you described it you end up with 461 (which makes sense because with x=5 you get 481 so x>5 is a must.) I think the way you calculated this is by treating this like normal brackets, in this case you would get x=4,78...
sweet editing
My thoughts as well!
That is so disappointing.
My bad, did not note it was a floor function prob. Should have known the problem would not be easy if you were taking the time to present it! At least we know answer is x=19/4 without the floor notation
Nice zoom
floor(x+1/27)=floor(x), floor(x+27/27)=floor(x+1). then we get, 78floor(x)+26+54=500, so floor(x)=16.15 implies contradiction
huh, that's wrong on multiple counts... First, floor( 0.99 + 1/27 ) is NOT equal to floor( 0.99 ). Then, there are 80 terms in the list, not 78. :-B
Good one; my approach was slightly different, as this term just begged to be completed :)
I add floor(x) to both sides and get:
floor(x) + floor(x+1/27) + ... + floor(x+80/27) = 500 + floor(x)
Then I group them in three of equal fractional part and get:
3 * (floor(x) + floor(x+1/27) + ... + floor(x+26/27)) + (1 + 2) * 27 = 500 + floor(x)
Hm. What is this sum of floors? Obviously it's actually floor(27 x)!
3 * floor(27 x) = 419 + floor(x)
Substitute y = 27 x.
3 * floor(y) - floor(y / 27) = 419
"Interesting". Whenever y crosses a boundary of integers, the LHS term increases by 2 or 3. There is no other case when the LHS term increases, and it's also obviously monotonous. This makes it excellent for guessing. Well, y=141 yields 418 < 419 and y=142 yields 421 > 419, so there is no solution.
How did I know to guess around y=141? Well, I ignored the floor and solved for y, and got 141.4125.
x= 4.75 (if it is a normal equation)
X= 5.17 (if it is a greatest integer fn)
These black bars are scaring me
Hey!
I was looking at it and thought about this:
Can you sum all X (i assumed that there is 80X because there is 80 terms) and the fractions but doint it the first fraction with the last getting 40 terms of (81/27) having:
80x + (1+80)/27 + (2+79)/27...
80x + (40)(81/27) = 500
80x + (40)(3) = 500
80x = 380
x = 38/8 = 4.75
Or there is a error on my process
This would be correct, if those were regular brackets. But those brackets represent the floor function, and you can't just add what's inside them together like you did. For example, [3.5]+[2.5]=3+2=5, and NOT 3.5+2.5=6.
Floor funtion? I didn't get that part. I havent seen the vídeo, just came ti check the result: 4,75
x = function() { 2... ? 0 : 500 }
may not be the kind of "x" that was asked for, tbf...
502 works. For example, 80*5+3*7 = 502 - 81 = 421.
That's an engineer's interpretation of "works"... Mathematicians wouldn't accept such an approximation. ;-)
I love the content!
Can we solve it by Arithmetic progression we know the value of a (first term) and we can find common difference which is 1/27 and we know the number of terms and then we also know the sum of Arithmetic progression Sn=n/2(2a+(n-1) d. Pls answer. Thx
Assuming you mean 80x + ((81*80)/2))/27 = 500? yeah that would be the simpler way to solve the problem
@@chrisbro3771 what's that did you use arithmetic progression
yea i keep getting 499 & 502 for some value of x never exactly 500
i added floor(x) + floor(x+81/27) [cuz 81/27 = 3] for completion/niceness sake (and fun) and u can get 500
Closest could be x~5.259259... (and it gives 499). But no solution that gives 500. Graphing could show this function rises with step of 3 (some cases with step of 1 or 2) and if step value is not solution, then no solutions.
What should this be? Primefinder?
Knowing 80n+3m =419 and 0
n=1 (mod 3) means n=1 or 4, not 1 or 5.
@@megalomorph thank . I edited
Thanks Sir.
if floor(x)=x we have a solution, x=419/80
I didn't noticed the floor operation, solved it and I got 15,5. Then I went to the end of the video and found "no solution" and got really confused xD
As you, I didn't notice the floor operation (parenthesis and floor sign are so close looking).
But for X I get a different solution from yours . Are you sure that X=15.5 is correct ? I get 19/4.
@@cret859 Yes, I made a mistake because I did it in bed a bit sleepy.
I did this:
There are 80 elements, each one is x + a/27, with a from 1 to 80. You can make 40 pairs that sum 2x+81/27 (you take the 1/27 and the 80/27, up to 40/27 + 41/27). That is 40(2x +3) = 500.
Then is simple
2x+3 = 12.5
2x = 15.5 (here I was, but I forgot the 2x)
x = 7.75
BUT that is not considering the floor operation xD
@@erik19borgnia Awake up Erik and check your answer, 40*(2*7.75+3) is not egal to 500.
I feel like professor Penn didn't argue convincingly about the domain for x. Why can't x be any number?
suppose x is positive, all the terms are grader than x, so x can't be greater than 500/80. One can reason the same way for x negative (notice that the floor is of -3.5 is not -3 but -4). This imply that the solution is restricted. I don't know if this helps or if I misunderstood your question. Best regards,
@@jaimeduncan6167 Thanks!
1/27+...+80/27 = 81x80/54=120. floor(1/27)+...+floor(80/27)=26x0+27x1+27x2=81.
That's 39 difference.
39=0(mod 3). It means 500 - 80.floor(x) = 0(mod3).
But x is between 5 and 6 (because 5x80+81=481, 6x80+81=561). So floor(x)=5. So there are no soutions.
Took me roughly 1 minute mentally. It's the exact same thing as Binary Agenda's reasoning, in fact, just expressed a bit differently. And all in all it s the same thing Michael Penn does in the video, but in a much more analytical and annoying way :).
The 500 - 80.floor(x) = 0 (mod 3) is why Michael found n=floor(x)=4 as a possible solution (but with m too large).
In the end you can work mod 80 and don’t have to go through all the cases
Or to reduce the number of cases further, use mod 3
It doesn’t mean there isn’t a solution. It means X doesn’t belong to [n ; n+1)
Hi Michael. I' m italian and follow your lessons with interest. Could you solve this problem ? Find all integer solutions of the equation : ax^2+5x+2=0 . Thanks for your courtesy .
The discriminant needs to be gte zero if there are to be any real solutions, so 25 - 8a >= 0. So checking on the first few values, we get (a=3, x=-1) ; (a=2, x=-2) ; (a=-3, x=2) ; (a=-7, x=1). And these are apparently all, at least when i interpret the question as saying that a is an integer as well, not just x. If that's not the case, we get infinite many solutions, including: (a=13/9, x=-3) ; (a=9/8, x=-4) ; (a=23/25, x=-5) ; (a=7/9, x=-6) ; etc.
@@irrelevant_noob thanks . This Is the same result that i obtain with wolfram alfa but i wanted a proceedings or the operation that i must do for have the result . Or a generalization of calculus . Ok thanks
@@irrelevant_noob only over the integer
@@fabiopompei317 no calculus needed, just a bit of algebra: to get x=n we need ... [ plugging in the quadratic formula: -5 +/- sqrt ( D ) / 2a = n , so +/- sqrt( D ) = 2an + 5 and squaring ... so: ] a = -(5n + 2) / n^2. Guess that was over-complicating things, could've gotten this just by plugging in x=n into the original ax^2+5x+2. ^^
PS "only over" WHICH integer? o.O
floor(x+(an-b)/n) where we take a,b,n,x as integers >0, the result will be x+a-1. Quickly we can figure that the sum will be 26(x+0)+27(x+1)+27(x+2)=500, simplify, 80x+81=500. We end up with x = 419/80.
But 80 ł 419 as a matter of fact 419 is a prime so no integer solutions
unsettling fact that floor(0.9999...)=1
um... sorry what? What are you doing, man? X is equal to 419/80. You're welcome.
question: is the floor of 3.999... (repeating) equal to 4?
No, the floor value of x is equal to the greatest integer lower than or equal to x. Even if x is very very close to 4, its floor value will be 3. And for negative numbers, that's the same thing ; so the floor value of -3.2 for example is -4, 'cause -4 ≤ -3.2
@@titou7367 but 3.999... (repeating) _equals_ 4 :S
Yes, the floor of 4 is 4
For those who thought like me "can't we write it as 80x +(81*40)/27=500 ?". the answer is -What is floor functions?
not that i know what it is either. :(
1. The thumbnail shows groupings of two elements, NOT groups of floors! So:
80x+(80⋅81)/(2⋅27)=500 →
80x+120=500 →
80x=380 →
x=4.75
Note: I have NEVER seen floor written so sides are connected underneath.
2. Say they are groups of floors. So:
26(x+0)+27(x+1)+27(x+2)=500 →
80x+27+54=500 →
80x=419 →
x=5.2375
Oh... I see...
JavaScript:
sum=0;
x=419/80;
for(i=1;i
1 is a trivial problem, and is not the subject of the video.
@@gurrrn1102
But that is what we see, not floor...
Channel owner should change it to two separate floor braces(?).
@@OrenLikes probably best not to assume an unorthodox notation has a common interpretation
@@gurrrn1102
If this is directed at me -
probably best not to speak when you have nothing intelligent to add.
If this is directed at the channel owner - you are correct!
@@OrenLikes it applies equally to you both.
this was moderate
Lost me at the third row.
19/4. Can anyone explain please how am I wrong?
floor function
Floor(x)=19/4 has no solution since Floor(x) must be an integer.
4.75 without floor nonsense
5.259 repeating gets close…
5.2375 is my final answer
The bloody hell is M doing in here?
awesome !!!