OK, let's kill it a bit fast by avoiding the Algebraic Geometry confusion. Suppose that x, y and z are the roots of a 3rd degree polynomial P(X) = (X - x)(X - y)(X - z) = X^3 - 62X^2 + cX - 2880. Since x, y and z are all greater or equal to 4, we have that P(4) =< 0. Plugging this in the monomial expansion of P, we obtain c =< 952. Since x, y and z do not exceed 40, we have that P(40) >= 0 and so, it follows that c >= 952. Thus, c = 952 and so, both of P(4) = 0 and P(40) = 0 should hold. Thus, 2 of the zeros (which are x, y and z) are 4 and 40, which means that the remaining one is 18, because their sum equals 62.
I think that the exercise is not about natural numbers and factorization, but about maximization/minimization of real functions. The only real solution with the three numbers in the interval is the one with one number at the upper border, another one at the lower border and the third one being the only number accomplishing the required condition of the sum and/or the product. If we fix the sum required (62), it is easy to verify that increasing the lowest real number (4) or decreasing the highest one (40) results in products higher than the one required (2880). On the other hand, if we fix the product required, it is easy to verify that increasing the lowest number or decreasing the highest one results in sums lower than the one required.
After proving that x, y, z are even, the problem simplifies to: find a, b, c in [2,20] s.t. a+b+c=31 and abc=360, which you solve observing that 31 is odd and considering a few options...
Yes, that's how I did it. Essentially equivalent to his mod 4 arguments, but easier to follow. After that, I just checked the 6 factors of 45 to be "C", then looked for solutions for a and b as factors of what was left; much easier and covered cases I think Dr. Penn left behind.
I really like how "guessing" the (4,18,40) solution can quite naturally follow up from the construction of the problem and trying to tackle it in multiple different ways here, even without trying to solve it lime that :-) One can just try having the smallest and biggest number from the [4,40] interval at the same time, then notice that 18=62-4-40 actually works and the product of x,y,z is 2880. One can try factorization and similar method to the one in the video, to put more work into it, but be sure to get all integer solutions. But even without it, if you simply try to just express any of the x,y,z using the other two, for example x=62-y-z, then because of x being in [4,40] interval, it gives us: 4
There are more real solutions outside the given interval, that's why we need that restriction: For example if we consider the case y = z, then we get x + 2y = 62 and xy² = 2880, which we can combine to get (62 - 2y)y² = 2880 or (31 - y)y² = 1440. Now we know that (31 - y)y² has its zeros at y = 0 and y = 31 and goes to infinity as y goes to negative infinity, so there is a solution with a negative y (and z) and a positive x. We can also calculate a local maximum at y = 62/3 with a value of (31 - 62/3)(62/3)² which is greater than 1440, so by the intermediate value theorem (31 - y)y² = 1440 must have a solution in (0, 62/3) and one in (62/3, 31). We can get better bounding intervals for them (like (7, 8) and (29, 30)) to show that they give us an x that is outside the interval [4, 40], but without that restriction they would constitute a second and third solution (plus permutations).
@@jesusthroughmary At :58 he says "we'll look for maybe natural number solutions first..." so it sounds like he's looking for all real solutions but is only considering integer solutions initially. I wonder which "famous theorem in algebraic geometry" he's referencing at 14:27 that gives an upper bound on the number of solutions.
You are right. Consider, for example, the following three numbers x = 8 y = 27 - 3 * sqrt ( 41 ) z = 27 + 3 * sqrt ( 41 ) It is straightfoward to see that x + y + z = 62 x * y * z = 2880 However, z > 40, so they are not in the interval. And, of course, y and z are not integers.
You are trying to use Bezout's Theorem at the end, but it does to apply in this case. In order to conclude that the maximum number of solutions is equal to the product of the degrees, you must have the same number of equations and variables. Here you have 2 equations and 3 variables. So actually, algebraic geometry says that there are infinitely many (complex) solutions. So you have not proven that there are no more solutions in [4,40]
I'd say x,y,z∈[4,40] eliminates complex solutions, especially when he read it as "on the interval". In any case at 1:00 he narrowed it to natural numbers.
Here's a solution that covers the real case: If we set k := xy + yz + zx, then x, y, and z are the roots of r^3 - 62r^2 + kr - 2880 = 0, or rearranged, r^3 - 62r^2 - 2880 = -kr. If we imagine k ranging across the real numbers, then possible solutions x, y, and z are given by the intersections between the cubic f(r) := r^3 - 62r^2 - 2880 and the line that passes through the origin g(r) := -kr. We need that all three of these intersections are in the interval [4,40]. We'll use this interval to give us restrictions on k. Looking at the left endpoint of the interval, f(4) = g(4) when k = 952. Since f'(4) = 3*4^2 - 124*4 = -448 > -952, the graph of f is less steep than the graph of g at r=4, so increasing k will decrease the coordinate of the intersection and vice versa. As such, k must be less than or equal to 952, or else one of our solutions will be less than 4. (This part of the proof is a little tricky to visualize.) In a similar fashion, on the right endpoint, f(40) = g(40) when k = 952 also. Since f'(40) = -160 > -952, we again get that increasing k will decrease the coordinate of the intersection and vice versa. This time, though, since the coordinate of the intersection must be less than or equal to 40, we get that k must be greater than or equal to 952. Combining these two, we have that k = 952, so our only solution is {x,y,z} = {4,40,18}. If we had gotten different boundary values of k using our left and right endpoints, we would have had a range of values for k, each producing a triple of values for x, y, and z. The other restriction that could have come up is that the line has to intersect the graph in three points (counting multiplicity). This puts a bound on the minimum and maximum values of k, corresponding to two values of k for which the line is tangent to f(r). We can find these values by solving the simultaneous system of equations f(r) = g(r), f'(r) = g'(r) for k and r. We'll get a negative solution for k and two positive solutions, and the range of possible values for k is between the two positive solutions (and also less than the negative solution).
good hello i'm brazilian so i don't know how to speak english but i'm using the translator, i wanted to say that i like your videos very much as i don't understand much but i really like math
If f(x)=x^3-62x^2+C*x-2880, then f(40)=10f(4). But we need the 3 roots of f to be within the interval [4, 40] so f(4) and f(40) must have opposite signs. So they must be equal to 0, and the roots are 4, 40 and hence 18. Thus this is the only real solution.
Is there an easy argument of why f(4) and f(40) have to have different signs? I think it's clear by looking at a graph of a cubic, but I don't like the "make a graph" argument since doesn't seem too formal.
The part at the end about there being only integer solutions is super wrong, there are infinite number of possible real solutions (just take x as constant and solve a simple system of equations). The mentioned theorem only works if the number of equations is not less than the number of variables (I think).
There are definitely an infinite number of solutions. We're looking for roots of the polynomial x^3 -62x^2 + Cx - 2880 =0, then by vieta's formulas the roots satisfy the equations we want. This polynomial has infinite solutions over the reals as we vary C, however only one of them lies in the given interval. The question has a unique solution, but only because x,y and z are bounded.
Wait i think your final argument with algebra theorem is wrong in general but can be salvaged with an other theorem in this specific case. Consider x < y < z in [a, b] (a>0) with a fixed sum S. Then the minimum of xyz is achieved iff two of them are on the boundary because concavity of logarithm. (otherwise take (x-eps, y , z+eps) : it will reduce the product and keep the sum intact) Therefore it's enough to check only solution when two of (x,y,z) are on the boundary thus x=4, z=40 and y=18 is the only solution
Ok, throwing in some algebraic geometry: The ideal/variety defined by the two equations is actually one dimensional, which means that there are infinitely many (complex) solutions. Once we fix one of the variables (e.g. x=a), it will either be unsolvable (x=0) or have dimension 0, in which case we can apply Bézout's theorem guaranteeing that there is exactly 3 projective complex solutions (counting multiplicities). Working in homogenous coordinates [x:y:z:w], homogenizing the equations leads to x+y+z-62w=0 xyz-2880w^3=0 x-a*w=0 Now, if w=0 (i.e. at infinity), we get x+y+z=0 xyz=0 x=0, so exactly the point [0:1:-1:0]. If a=0, from xyz=0, we actually get that this is a triple root, so there is no affine solution. if w!=0, we dehomogenize w=1 and get to the affine system we wanted to solve. It will now have 2 further solutions (unless a=0), and those are actually easy to find: Using x=a and rearranging the first equation gives y=62-a-z Inserting this into the second equation, we get a*(62-a-z)z-2880=0 hence -az^2+a*(62-a)z-2880=0. This is simply a quadratic equation in z. It's roots are real, if the discriminant D=a^2(62-a)^2-4*2880a is nonnegative (in the case D=0, z=y is a double root) and it is rational, if D happens to be a square of a rational number. Since D is a quartic polynomial with leading coefficient 1, for big and small values of a, the roots will be real. The only remaining question is, for which rational a this is a rational number. Writing a as b/c and clearing denominators, we get Dc^4=b^2(62c-b)^2-11520c^3b. The RHS is an integer, the LHS by assumption a square of a rational number. This means that Dc^4 actually already is a perfect integer square. Hence, we're looking for integer squares of the form b^2(62c-b)^2-11520bc with coprime c and b. Note that b is a divisor of the RHS, so it must also be a divisor of the LHS. But then, b^2|LHS, hence b^2|RHS-b^2(62-b)^2=-11520bc^3. Dividing by b yields b|-11520c^3. Since b and c are coprime, b|11520. These are only finitely many cases (divisors of 11520), and I think there aren't too many possibilities for c in each case. Going through these will finally settle the question about rational solutions.
Suggested question (I want to learn if this is a known theory in mathematics): 1- If we have an indefinite integral, is there a method to prove that the anti-derivative is expressible/non-expressable in elementary functions? 2- If we have a definite integral, is there a method to prove that it's value is expressible/non-expressable in constants like π, e, √2,... or rational numbers? Just hit like if you are willing to discuss this question.
similarly, at around 5:30, shouldn't we consider the possibility that none of x,y,z are congruent to 2 (mod 4)? for instance if two of them are congruent to 1 (mod 4) and the last is a multiple of 4, that would work -- except that it would require the last number to be bigger than 62
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@@tylerbreisacher5841 We already proved that all of them have to be even at this point. So 1 (mod 4) no longer applies.
But that possibility has already been proven wrong and it’s trivial. The product of odd numbers would never give an even answer. I agree that he should have included it but it doesn’t break the world lol
Why not check them all... once you know x is 2 mod 4... eliminate z from the 2 equations to get one quadratic in terms of x and y. Solving for y, the discriminant is x^4-124x^3+3844x^2-11520x = x^2(x^2-124x+3844-11520/x). Possible values of x are 2,6,10,18,30,90. We can eliminate 90 because the sum is 62. Substitute the others into x^2-124x+3844-11520/x and see which are perfect squares. 2 gives a negative discriminant. The other values are (x,discriminant) are (6,1216), (10,1552),(18,1296),(30,640). The only perfect square is 1296.
You can't disregard the bounds for x,y and z, they make the solution unique. Any triple of roots to the cubic x^3 - 62x^2 + Cx - 2880 (C is a free variable) is valid solution, the solution you found is the only one which lies in the interval [4, 40].
@@angelmendez-rivera351 I know I know, I was referring to his comment near the start. He said that he would find all solutions by finding the solutions on [4, 40], which is not the case.
Here are some viewer suggested problems, from my Oxford interview about 20 years ago! (i) if you start at the bottom left corner of an m by n chessboard, how many unique paths are there to the top right corner if you can only move one step right or one step up each go? (ii) how many zeros does 100! have? A lot easier than the Olympiad questions but only had a few minutes to figure them out (but with tutors happy to give hints in places - part of the interview was checking they could actually teach you!)
What if.. we set up the cubic equation t^3 - 62t^2 + mt - 2880 =0, where x, y, and z are its roots (real), and discussed according to the values of the real parameter m their existence and then computed them in that way? Maybe include a little calculus (variations of a the cubic function) in that...
@@CauchyIntegralFormula nicely done my friend.... i rarely watch at my desk, and so was too lazy to get up and give it a try. Sorry I missed your comment bud.
I don't see where the problem says to take x,y,z to be integers. But if indeed they are, I don't understand why you didn't just do the case analysis. Given x,y,z in [4,40], then by the definitions of a,b,c, we have a in {3,5,7,9,11,13,15,17,19}, b in {1,3,5,7,9}, c in {1,3,5}. Then using the fact that abc = 45, we can further restrict to divisors of 45, giving a in {3,5,9,15}, b in {1,3,5,9}, c in {1,3,5}. Then if a = 15, bc = 3, so (b,c) is one of (1,3),(3,1), neither of which satisfy b+2c = 8. If a = 9, bc = 5, so (b,c) is one of (1,5),(5,1). (5,1) doesn't satisfy b+2c=11, but (1,5) does, so we have a solution (a,b,c) = (9,1,5). If a = 5, bc = 9, so (b,c) is one of (3,3),(9,1), neither of which satisfy b+2c=13. Finally, if a = 3, bc = 15, so (b,c) is one of (3,5),(5,3), neither of which satisfy b+2c = 14. So the only solution to a+2b+4c = 31 and abc = 45, with a,b,c odd integers and 2a,4b,8c in [4,40] is (a,b,c) = (9,1,5).
Just so you know, I've made 100's of videos for my math classes, and ending the video has always felt awkward. I'm stealing, "and that's a good place to stop."
Problem suggestion, comes from one of the BMO's. N is a four-digit integer, not ending in zero, and R(N) is the four-digit integer obtained by reversing the digits of N; for example, R(3275) = 5723. Determine all such integers N for which R(N) = 4N + 3.
it would be so much better if the problem had a xy+yz+xz=something because that would imply the 3 symmetric sums for vietas formulas on a 3rd degree polynomial.
Hmm, would you not need to prove that x, y, and z have to be integers to say that "x, y, and z must be even"? The interval [4, 40] contains way more than just N \cap [4, 40], after all.
Well if I got it right he kind of assumes that the solutions would be from Integers and eventually shows that the 6 solutions he found are all the possible ones because of a theorem (I honestly don't remember or even know) on polynomial equations...
We are just looking for integer solutions first. So You just need to add: if x,y and z are all integers, then they must ... The final arguments says that this type of problem has 6 solutions max, and as we already we found 6 integer solutions, all solutions are with integers.
Prove that every natural number can be written as the sum of numbers of form 2^m . 3^n where no terms divide each other. E.g. 23 = 9 + 8 + 6 49 = 27 + 18 + 4 This was the first question on a Putnam exam but I can’t find a video.
without the bounds, there are infinitely many solutions, which you can see by eliminating z and feeding the result into wolfram alpha (which produces a fascinating graph like a three-way parabola with a curved triangle in the middle). the clever part about the bounds is that they cut out a square subset of this graph where a vertical (or horizontal) line will pass through it twice at only a single point - x or y = 18. either way, you necessarily get the other two values at the bounds of 4 and 40. if the solution /didn't/ contain 4 and 40, there would be other solutions in the reals. so with a conspicuous setup like this, a good first guess might be to try the bounds as two of the values and see if that produces a solution. alternatively: if you fix y = n, you get a quadratic x²n + x(n² - 62n) + 2880 = 0 where the two roots are the values for x and z. but those roots must be within [4, 40], so the question is how to guarantee that. a reasonable guess is to set the midpoint of the parabola at the midpoint of the interval, so (62 - n) / 2 = (4 + 40) / 2 = 22, which gives n = 18 and yields the solutions again. unfortunately i'm not sure how to prove this is the only possibility, since the discriminant ends up being a pretty gnarly quartic.
Goodness! I just listed the factors of 2880 (4,5,6,78,9,10,...,40) and went through the possible sets from (4,4,180) to (12,15,16). And the only one that summed to 62 was (4,18,40).
Wow, with these types of problems I'm usually like "well I'll just find one solution (and by symmetry 6 in this case)". No idea I was pretty much done then
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Well, you are done finding solutions. But you are not done proving that there are none or no more.
If you wish to show that apart from the ones we have shown, there are none other real solutions all in the interval [4,40], you can do as follows: 1. Show that there doesn't exist a solution in the interval such that x=y (or any other permutation) 2. Define F(x,y,z)=(x+y+z-62,xyz-2880). Suppose there exists a solution (x0,y0,z0) in (4,40)^3, and WLOG x0
Pretty sure it was just experience, since we had all numbers even numbers, and because 2880 only had 3 prime factors our only real choices were 4,3,9,5.
No need to get idea that is mod 4 or not Noted that x,y,z are all even, Ignore the odd factors the combinations for x,y,z are 1.[2a, (2^2)b, (2^3)c] 2. [(2^2)a, (2^2)b, (2^2)c] 3. [(2^4)a, (2)b, (2)c] the mod 4 test can quickly reject the combination of [2 and 3]
The arguments about x,y,z (mod 4) imply, WLOG, x=4a+2, y=4b, z=4c. The original equations then reduce to: (2a+1)bc = 90 and (2a+1)+2(b+c) = 31. There are only 10 distinct factorizations of 90 and it's straightforward to check, for each one, whether (odd factor)+2x(other 2 factors) = 31; only (9,1,10) qualifies, giving the solution (18,4,40).
Here's my solution over the positive integers: The prime factorization of 2880=2^6×3^2×5. Because the sum is even, either all summands are even, or one of them is. But if only one of them is, it must contain all 6 factors of 2, so it must be at least 64, which is too large. So they are all even so that we can write 2a + 2b + 2c = 62 and 2a2b2c=2880 Or a+b+c=31 and abc=360 Ok, great Now a,b,c cannot be all odd since abc is even, so one of abc must be odd. There are 2 factors of 2 left, so solutions must have the form a=e, b=2f, c=4g where e,f,g odd. Let's see what that gives us. e+2f+4g=31 efg=45=3^2×5 Fantastic (play with chalk) g must be smaller than 31/4, so it is either 1, 3 or 5 Case 1. g = 1. We have e+2f=31-4=27, ef=45 1×45 the sum 2+2f is either 47 or 91, both too large 3×15: sum is 21 or 33 5×9: sum is 19 or 23 So case 1 does not give us any solution. Nice. Lets see what case 2 gives us. Case 2. g = 3. We have e+2f=31-12=19, ef = 45/3=15 1×15: sum is 16 or 31 3×5: sum is 13 or 11 No solution for case 2 either. Great. Finalle move to case 3. Now let's erase the board (backflip). Case 3. g = 5. We have e+2f=31-20=11, ef = 45/5=9 1×9: sum is 11. With e=9 and f=1 the e+2f=11 so this is a solution. Plugging this in we get 2a=2e=18, 2b=4f=4, 2c=8g=40 18+4+40=62 and also 18×4×40=2880. Perfect, this is a solution. With e=1 f=9 the result of e+2f is 19. Setting both e and f to 3 gives e+2f=9, fo we have no further solutions and this is a good place to stop.
Hmm I proved that all of x, y, z have to be even. Then for one of them (lets say x) it has to be: 10|x, so x in {10, 20, 30, 40} and then you can check easily for possible solutions. For example: X=40 => y=22-z => 22z-z²=72 => z²-22z+72=0 => z=11+/- sqrt(11²-72) => z=4 v z=18
Without the interval restriction on values, the solution is the intersection of the two 3-D curves given by the equations in x,y,z. One curve is a plane and the other is a set of hyperbolic shapes. The locus of solutions includes, for example, (3.5, 23.5311402, 34.9688598). Another solution is (-2, -17.64, 81.64) roughly. It would be nice to redo the problem and show the shapes involved and the interval restriction which touches the solution locus at only six points. The fact that there are only six solutions has nothing to do with number theory and everything to do with the clever construction of the problem.
This is a Floor Function problem from the Indian National Maths Olympiad 2014 problem 2 (since you love floor functions so much) The problem states : Let n be a natural number , prove that floor(n/1) + floor(n/2) + floor(n/3) + ... + floor(n/n-1) + floor(n/n) + floor(sqrt(n)) is even , below is the link to official pdf of the problem olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/inmo-2014.pdf Thanks
X=40 or 18 or 4, Y=18 or 4 or 40, Z=4 or 40 or 18. Easy decision. This is not a hard problem at all. Just plug in 4 and 40 to either xz and solve for y, to xy and solve for z, or to yz and solve for x.
If you disregard the restriction x,y,z є [4;40] there are tons of solutions, if you substitute z for 62-x-y in the second equation you get a third order curve: imgur.com/a/ZoawDle . It is only due to x,y being forced to be in a hexagon by the restrictions 4
From the fact that x,y,z are all even, one may write x=2x', y=2y', z=2z' so that x'+y'+z'=31 and x'y'z'=360. The first equality implies that exactly one, or all, are odd. However, the second equality excludes the second alternative. So one is odd. There is only 3 possibilities in the interval [2,20] : 5, 9, 15. This may help a little bit.
6:35 "So if we get a single solution, that's actually 6 solutions…" Assuming the 3 numbers are distinct. If the numbers were (A, A, B), there are only 3 distinct permutations. For the record, brute force with a spreadsheet solved this faster than this entire video. Sometimes, brute force is the best way.
I think the brute force only works when assuming integer/natural number solutions - but that is what I did too, but with T-SQL instead: with a as (select * from dbo.fn_sequence(4,40)) select * from a a cross join a b cross join a c where a.id+b.id+c.id=62 and a.id*b.id*c.id=2880 where fn_sequence is a function that produces a table of integers in the given range.
Wow, he made this too complex. Recognize that x+y+z=62 is the equation of a spherical shell. So, you have a limited number of solutions. Because x, y, and z are all factors of 2880; x, y and z must be multiples of the base factors or 2, 3, and 5. The interval of 4 to 40 limits your possible multiples for 2 and 3. The power of 3 must be 2 because the multiple must be greater than or equal to 4. The multiple of 2 can be 2, 3, 4, or 5 times. 1 times is less than 4 and 6 times is greater than 40. You can also eliminate 5 as 2880 only has one factor of 5 and that is already accounted for. So, you have to test 2, 3, and 4 powers of 2 for the last number.
I have found all solutions to the system of equations x + y + z = 62 and x * y * z = 2880: (x, y, z) = (31-z/2+K, 31-z/2-K, z), where K = sqrt[(31-z/2)^2 - (2880/z)] = sqrt[z^3 - 124*z^2 + 3844*z - 11520] / [2*sqrt(z)]. Note that if we solve for K = 0, then we get x = y = 31 - z/2 and z = 1 of the roots of z^3 - 124*z^2 + 3844*z - 11520.
I have come to think that the problem was intended over the real numbers, but that Michael just assumed it was over the integers. And that his remark about natural numbers was only intended as a restriction to positive integers. And it is a much more interesting problem over the real numbers! For real x,y,z, having xyz=2880 and x+y+z=62 we could pick any value from the given interval for x, and then set yz=2880/x and y+z=62-x. All solutions then have y^2+z^2 = (y+z)^2- 2yz = (62-x)^2-5760/x. This corresponds to a circle in the plane with constant x (parallel to the y-z-plane) with radius ranging from sqrt(1940) at x=4 to sqrt(340) at x=40. (The minimum radius occurs at x>62, so well outside the range). For any choice of x in [4, 40] the circle intersects the square region having y, z in [4, 40] and it should also intersect the line y+z=62-x. When we pick an x0 with help of Wolfram Alpha, I found the solutions when y
There are going to be other real solutions without the restriction to [4,40]. Your equations reduce to x + y = 62 - z and xy = 2880/z, so x and y are the roots to r^2 + (z - 62)r + 2880/z = 0. The two roots will be real if (z- 62)^2 > 4*2880/z or z(z -62)^2 > 4880. This clearly can happen, so there are definitely other triples of real solutions. You'd need an additional argument to show they can't all be in [4,40].
awesome! @1:00 I paused and factored 2880 to (2^2,2^3*5,whats^left) and they summed to 62........ yea, inside the interval notation.......the FBI calls that a clue as well
An Elemantary way.. Since all number are even and there is only one 5...so one of the number must be 10, 20,30 or 40... Forming equations and putting values of z as 10,20,30, 40 we can get values of other numbers
Solving the system: x+y+z = 62 x*y*z = 2880 gives a one-parameter solution. Taking z as the parameter, and assuming x 40, so these values of z do not give a solution within the required interval. Similarly, for z in (18, 40), x < 4, so these also do not give a solution within the required interval. Only by assigning z a value from the set { 4, 18, 40 } are both x and y also in [ 4, 40 ], so there are indeed only 6 solutions, all of them being some permutation of (x, y, z) = (4, 18, 40). However, if the restriction to the interval [ 4, 40 ] is lifted, then every value of z such that (z - 62)^2 >= 11520/z leads to real values for x and y, and these (x, y, z) values are all solutions to the system of equations. So, there are, in fact, an uncountably infinite number of solutions, if x, y, and z are allowed to be any real numbers.
no! @Micheal Penn, the interval has a meaning. And why are you considering x,y,z to be naturals?? It would have been a lot easier if you've considered reals. A similar problem on geometry I created, used this problem application. Do have a look at this. Problem: artofproblemsolving.com/community/c1619209h2357485_geometry_problem__1 Solution: artofproblemsolving.com/community/c1619209h2357490_geometry_problem__1solution
I found that this solution was the only one with integers, but I remainded stun by your proof that there are no other real solution : so obvious ... but only when you 've said it !
In fact, he's wrong. There are not just 6 real solutions, the intersection of two surfaces in R^3 is a curve, so there are infinitely many real solutions.
@Angel Mendez-Rivera I mean, his solution/logic is incorrect. There are in fact 6 solutions, but the proof given is wrong. His proof was " Find 6 integer solutions and said that there are only 6 because the max number of real solutions is 2*3=6" That is completely wrong.
I got the answer by comparing (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc to x^3 - 62x^2 + Lx - 2880. By setting c = 40, it was a quadratic problem to show a = 4 and b = 18, so L = 952. If L (=ab + ac + bc) equals anything other than 952, then a or c fall outside the bounds. Hard to prove why though.
To clarify: I graphed y = x^3 - 62x^2 + Lx - 2880 in Desmos. After finding that L = 952 by hand, it's clear that any other value of L pushes at least one zero outside the range [4,40]
I do it see above. In fact the unique solution over the naturals is x=4,... And in general. I have x+y+z=62, xyz=2880 then x+y=62-z and x*y=2880/z then 2*x=(62-z)+-√((62-z)^2-4*2880/z). Fix z then there is solution over R iff (62-z)^2-4*2880/z>=0. If z>0 then there is solution iff z*(62-z)^2-4*2880>=0. And if x,y,z>=0 then there is solution iff z*(62-z)^2-4*2880>=0 and 62-z>=0,z
We have 0=f(x)=x^3-62*x^2+L*x-2880. But idea is that L=(2880+62*x^2-x^3)/x and graph L with function of x. We observe with a point a have 3 pre-imagens.
We also have the the discriminant of the polynomial is the form -4*(L^3+a*L^2+b*L+c) then the discriminant is positive over the limited region. If L in Z then the discriminant is a square in the finite points the region. This condition is necessary but not suficient for the roots are integers.
I am a truck driver, so as a matter of professional interest, I think it would be neat if you could derive the equation for a tractrix. It would be a little different from the usual geometry and number theory problems you solve.
Notice the multiset of prime factors of 2880 is {2,2,2,2,2,2,3,3,5}. (x,y,z) must exhaust this multiset. (x,y,z) ∈ [4, 40] -> (x,y,z) ∈ {4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 32, 36, 40} let (a,b,c) be the un-ordered version of (x, y, z)
let a contain exactly one of {5, 10, 15, 20, 30, 40} e.g., it contains the only 5 in the prime factors. If a=5, then b+c = 57. This means one of the two must be odd and the other even, therefore b=9 and c=38, but 38 is not in the range of possible values, so no (x,y,z) contains 5 as an element. If a=15, then b+c = 47. This means one of the two must be odd and the other even, therefore b=9 and c=28, but 28 is not in the range of possible values, so no (x,y,z) contains 15 as an element. If a=10, then b+c = 52. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (16, 36) is the only pair that sum to 52. 10=2*5, 16=2^4, 36=2^2*3^2, the product abc is 2^7 * 3^2 * 5 != 2880 so cannot be a solution. If a=20 then b+c = 42. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (6, 36) and (18, 24) both pairs contain 3 threes in the join of multiset their multisets so cannot be solutions. If a=30 then b+c = 32. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (8, 24)) is the only pair that sum to 32. 8=2^3, 24=2^3*3 the produce abc is 2^7*3*5 != 2880 so cannot be a solution. If a=40 then b+c = 22. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (4, 18) and (6,16). The product 6*16*40 != 2880 but 4*18*40 = 2880. This exhausts the possible values for a, therefore all six permutations of 4, 18, 40 are the only solutions to x+y+z=62, xyz=2880, and (x,y,z) in [4, 40]
It's the "unique" answer. But if x,y,z are roots the polynomial f(t)= t^3-62*t^2+r*t-2880. By Vieta's identities then r=(2880+62*u^2-u^3)/u. Where u=x,y or z. Then the function q(u)=(2880+62*u^2-u^3)/u take the value of x there times. Graph q(u) over interval [4,40] and verify the unique r satisfying this is q(4)=q(40).
www.wolframalpha.com/input/?i=%282880%2B62*u%5E2-u%5E3%29%2Fu+from+4++to+40. Graph of q(u). We can prove using methods of fundamental calculus to q(u) but is long.
Thanks for all your videos. The way you present the solution is interesting. I have one question for you. Prove that the number of digits in sum of all n digit natural numbers is 2n.
Looking for all integer valued solution would give the same answer. Here is the analysis: Step 1. Only one is even is impossible since it would imply |x|= 64, |yz|=45, y+z=-33 or 95. Thus, all are even. Also one of them is divisible by 5. Hence, without loss of generality, we write x=10a, y=2b, z=2c. This gives the equivalent system b+c=31-5a, abc= 8*9 Step2. The case a=y>=z) is x= 40, y=18, z=4.
@@dinkband7662 yes I know mod but it's application is hard. Just need to practice some questions, will get hang of it. It's surprising where it's application pops up.
Why are we limiting ourselves to integers? The closed interval [4,40] is not the same as the set {4,5,...,40}. And even the notation x,y,z is suggestive of reals as opposed to integers k,m,n for instance.
I was guessing right from the beginning and found the solution quickly. Some of the "2" factors has to be joined with 9 and 5. There are only a few ways you can do this. 5 can be combined with maximum 3 two's (else the sum is higher than 62). 9 can only be combined with 1 or 2 two's (else the sum is higher then 62). So, in the end you've to check only 6 possibilities.
These are one of those questions which u see and think "yaa I can do this" and only to realize later that u have missed many more solutions and u have got some values which don't even satisfy the system(at least happens to me quite often) :p
once you show all are even: let a,b,c = x/2,y/2,z/2. a+b+c = 62/2 = 31 abc = 2880/8 = 360 = 2^3 x3^2 x5 so the problem has been simplified. once again we ask about parity for a,b,c. 2 even: even+even+odd = odd is okay, and is the only okay combination. therefore we know (without loss of generality) a=4d, b=2e. c+4d+2e = 31 cde = 45 where c,d,e are all odd numbers (in the set 1, 3, 5, 9, 15, 45) we can eliminate 45 as an option, and begin enumerating. (1,3,15), (1,5,9), (3,3,5): the only three valid unordered triples. 15+12+2 = 29; 15+4+6 = 25; 3+4+30 = 37; 1+12+30 = 43: no solution with (1,3,15) 1+20+18 = 39; 5+4+18 = 27; 9+20+2=31: one solution with (9,5,1) 3+20+6 = 29; no solution with (3,3,5) therefore (x,y,z) in any order are (18,40,4)
I want to ask you. when I study from a textbook , should I solve all problems after every section or chapter (too many problems and need much time , also there 's 5 subjects or 6 every semester) or just select some problems which i feel are difficult or have new ideas .
There are actually infinitely many real solutions. I am not familiar with algebraic geometry so I don't know the theorem he is referring to here but I am pretty sure it does not apply to this situation. Note that x = 4, y = 18 and z = 40 are in the lattice [4, 40]^3. If x increases, then y and z decrease up until y (or z) reaches 4 at which point you get another permutation on the lattice. So there are infinitely many solutions in the interior of the cube. I watch the video to see if he found a nice representation of the solutions but ended up being disappointed by all the work he did to reach the solution on the lattice and a wrong conclusion to the problem.
Am I missing something: A system of two equations within three variables has a one-parameter family of solutions by the implicite function theorem. If these six solutions are the only solutions to the problem, it definitely must have something to do with the interval [4, 40].
@@angelmendez-rivera351 This is true, but any extension of the interval by an arbitrarily small amount would give infinitely many solutions. (Note that all points on the curve are in fact regular)
Bismillaah. Solve X+ Y+ Z = 62 XYZ = 2880. Solution. First Verify the Factors of 2880. Such that The Factors Should be three Only and Sum of Factors Should be Equal to = 62. Then.. The Factors are. Since 62 is Even. All the three Factors Should be Even Or Atleast One Factor Should be Even. 2880 is Even. Here also All the The Three factors Should be Even Or Atleast One Factor should be Even. And Factor should be Multiple of 5 Because Last Digit is 0.(Zero). Therefore. One o f three Factors should be Multiple of 5 Or 10. Some Factors of 2880. 1)80× 4×9.= 2880 But 80+4+9= 93. 2) 40× 8×9 = 2880 & 40 + 8+9 = 57 3) 40× 18× 4 = 2880 & 40 + 18 +4 = 62.. Or 4 + 18+ 40 = 62 It is the the Answer. Therefore X = 4. Y = 18 Z = 40
OK, let's kill it a bit fast by avoiding the Algebraic Geometry confusion.
Suppose that x, y and z are the roots of a 3rd degree polynomial P(X) = (X - x)(X - y)(X - z) = X^3 - 62X^2 + cX - 2880. Since x, y and z are all greater or equal to 4, we have that P(4) =< 0. Plugging this in the monomial expansion of P, we obtain c =< 952. Since x, y and z do not exceed 40, we have that P(40) >= 0 and so, it follows that c >= 952. Thus, c = 952 and so, both of P(4) = 0 and P(40) = 0 should hold. Thus, 2 of the zeros (which are x, y and z) are 4 and 40, which means that the remaining one is 18, because their sum equals 62.
I think that the exercise is not about natural numbers and factorization, but about maximization/minimization of real functions. The only real solution with the three numbers in the interval is the one with one number at the upper border, another one at the lower border and the third one being the only number accomplishing the required condition of the sum and/or the product. If we fix the sum required (62), it is easy to verify that increasing the lowest real number (4) or decreasing the highest one (40) results in products higher than the one required (2880). On the other hand, if we fix the product required, it is easy to verify that increasing the lowest number or decreasing the highest one results in sums lower than the one required.
After proving that x, y, z are even, the problem simplifies to: find a, b, c in [2,20] s.t. a+b+c=31 and abc=360, which you solve observing that 31 is odd and considering a few options...
Yes, that's how I did it. Essentially equivalent to his mod 4 arguments, but easier to follow. After that, I just checked the 6 factors of 45 to be "C", then looked for solutions for a and b as factors of what was left; much easier and covered cases I think Dr. Penn left behind.
Yes, exactly equivalent...but easier
I really like how "guessing" the (4,18,40) solution can quite naturally follow up from the construction of the problem and trying to tackle it in multiple different ways here, even without trying to solve it lime that :-)
One can just try having the smallest and biggest number from the [4,40] interval at the same time, then notice that 18=62-4-40 actually works and the product of x,y,z is 2880.
One can try factorization and similar method to the one in the video, to put more work into it, but be sure to get all integer solutions.
But even without it, if you simply try to just express any of the x,y,z using the other two, for example x=62-y-z, then because of x being in [4,40] interval, it gives us:
4
There are more real solutions outside the given interval, that's why we need that restriction: For example if we consider the case y = z, then we get x + 2y = 62 and xy² = 2880, which we can combine to get (62 - 2y)y² = 2880 or (31 - y)y² = 1440. Now we know that (31 - y)y² has its zeros at y = 0 and y = 31 and goes to infinity as y goes to negative infinity, so there is a solution with a negative y (and z) and a positive x. We can also calculate a local maximum at y = 62/3 with a value of (31 - 62/3)(62/3)² which is greater than 1440, so by the intermediate value theorem (31 - y)y² = 1440 must have a solution in (0, 62/3) and one in (62/3, 31). We can get better bounding intervals for them (like (7, 8) and (29, 30)) to show that they give us an x that is outside the interval [4, 40], but without that restriction they would constitute a second and third solution (plus permutations).
I think he meant that [4,40] yields the same set of solutions as the set of all natural numbers.
@@jesusthroughmary At :58 he says "we'll look for maybe natural number solutions first..." so it sounds like he's looking for all real solutions but is only considering integer solutions initially. I wonder which "famous theorem in algebraic geometry" he's referencing at 14:27 that gives an upper bound on the number of solutions.
@@Jaeghead good point
@@Jaeghead That's what I'd like to know too... it seems with too many degrees of freedom to allow for a finite number of solution!
You are right.
Consider, for example, the following three numbers
x = 8
y = 27 - 3 * sqrt ( 41 )
z = 27 + 3 * sqrt ( 41 )
It is straightfoward to see that
x + y + z = 62
x * y * z = 2880
However, z > 40, so they are not in the interval.
And, of course, y and z are not integers.
You are trying to use Bezout's Theorem at the end, but it does to apply in this case. In order to conclude that the maximum number of solutions is equal to the product of the degrees, you must have the same number of equations and variables. Here you have 2 equations and 3 variables. So actually, algebraic geometry says that there are infinitely many (complex) solutions. So you have not proven that there are no more solutions in [4,40]
I'd say x,y,z∈[4,40] eliminates complex solutions, especially when he read it as "on the interval". In any case at 1:00 he narrowed it to natural numbers.
Yeah, at the end the problem is mathematically is reduced to one equation with two unknowns.
@@onderozenc4470 how? he just tested one case with b=1
At 8:51 it should be 2^6 on both sides. Inconsequential mistake, abc is still equal to 3^2*5
No, he's actually divided the product by 2 already, since the addition above also was divided by two.
@@Antanana_Rivo he points to the equation xyz = 2880, and xyz = 2^6*abc. Again doesn't make a difference, just pointimg it out
Here's a solution that covers the real case:
If we set k := xy + yz + zx, then x, y, and z are the roots of r^3 - 62r^2 + kr - 2880 = 0, or rearranged, r^3 - 62r^2 - 2880 = -kr. If we imagine k ranging across the real numbers, then possible solutions x, y, and z are given by the intersections between the cubic f(r) := r^3 - 62r^2 - 2880 and the line that passes through the origin g(r) := -kr. We need that all three of these intersections are in the interval [4,40].
We'll use this interval to give us restrictions on k. Looking at the left endpoint of the interval, f(4) = g(4) when k = 952. Since f'(4) = 3*4^2 - 124*4 = -448 > -952, the graph of f is less steep than the graph of g at r=4, so increasing k will decrease the coordinate of the intersection and vice versa. As such, k must be less than or equal to 952, or else one of our solutions will be less than 4. (This part of the proof is a little tricky to visualize.) In a similar fashion, on the right endpoint, f(40) = g(40) when k = 952 also. Since f'(40) = -160 > -952, we again get that increasing k will decrease the coordinate of the intersection and vice versa. This time, though, since the coordinate of the intersection must be less than or equal to 40, we get that k must be greater than or equal to 952. Combining these two, we have that k = 952, so our only solution is {x,y,z} = {4,40,18}.
If we had gotten different boundary values of k using our left and right endpoints, we would have had a range of values for k, each producing a triple of values for x, y, and z. The other restriction that could have come up is that the line has to intersect the graph in three points (counting multiplicity). This puts a bound on the minimum and maximum values of k, corresponding to two values of k for which the line is tangent to f(r). We can find these values by solving the simultaneous system of equations f(r) = g(r), f'(r) = g'(r) for k and r. We'll get a negative solution for k and two positive solutions, and the range of possible values for k is between the two positive solutions (and also less than the negative solution).
found it!!
good hello i'm brazilian so i don't know how to speak english but i'm using the translator, i wanted to say that i like your videos very much as i don't understand much but i really like math
Just watch this impressive Math channel th-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
@@oilermschmaharasta2905.thank you friend
Esse canal é uma mina de ouro! Boa sorte com a matemática
14:43 Loading 99.1%...
Congratulations, IMO Busters!
Very disappointing ending
If f(x)=x^3-62x^2+C*x-2880, then f(40)=10f(4). But we need the 3 roots of f to be within the interval [4, 40] so f(4) and f(40) must have opposite signs. So they must be equal to 0, and the roots are 4, 40 and hence 18. Thus this is the only real solution.
Is there an easy argument of why f(4) and f(40) have to have different signs? I think it's clear by looking at a graph of a cubic, but I don't like the "make a graph" argument since doesn't seem too formal.
@@angelmendez-rivera351 Oh right, f(4)-infty as x->-infty) there would be a root in (-infty,4). Similarly f(40)>=0. This is what I wanted.
how do you know that f(40)=10f(4)
if f(x)=x, then f(40)=10f(4), but why in this case
@@johnurga275 Plugin and you'll see.
@Julian Mejia
ok
Find all pairs (a, b) of integers a, b ≥ 1 that satisfy the equation
a^b^2=b^a
This is from the 1997 IMO :)
Is that (a^b)^2 or a^(b^2)?
@@angelmendez-rivera351 Wouldn't that be reading from right to left?
I made a video solving that one :)
th-cam.com/video/5O5ZPuag2Yo/w-d-xo.html
Thank you very much. Can you be more specific about the nb of solution part? Does that cover real solutions as well?
Is more long. But I wrote a idea
Yes, this is unclear, as there are 3 variables, but only 2 equations, thus potentially an infinite number of solutions.
The part at the end about there being only integer solutions is super wrong, there are infinite number of possible real solutions (just take x as constant and solve a simple system of equations). The mentioned theorem only works if the number of equations is not less than the number of variables (I think).
There are definitely an infinite number of solutions. We're looking for roots of the polynomial x^3 -62x^2 + Cx - 2880 =0, then by vieta's formulas the roots satisfy the equations we want. This polynomial has infinite solutions over the reals as we vary C, however only one of them lies in the given interval. The question has a unique solution, but only because x,y and z are bounded.
@@angelmendez-rivera351 I was referring to the solution to the problem given, not the video. This response misses the point.
@@angelmendez-rivera351 ㅐ
Wait i think your final argument with algebra theorem is wrong in general but can be salvaged with an other theorem in this specific case.
Consider x < y < z in [a, b] (a>0) with a fixed sum S.
Then the minimum of xyz is achieved iff two of them are on the boundary because concavity of logarithm.
(otherwise take (x-eps, y , z+eps) : it will reduce the product and keep the sum intact)
Therefore it's enough to check only solution when two of (x,y,z) are on the boundary thus x=4, z=40 and y=18 is the only solution
Ok, throwing in some algebraic geometry: The ideal/variety defined by the two equations is actually one dimensional, which means that there are infinitely many (complex) solutions.
Once we fix one of the variables (e.g. x=a), it will either be unsolvable (x=0) or have dimension 0, in which case we can apply Bézout's theorem guaranteeing that there is exactly 3 projective complex solutions (counting multiplicities).
Working in homogenous coordinates [x:y:z:w], homogenizing the equations leads to
x+y+z-62w=0
xyz-2880w^3=0
x-a*w=0
Now, if w=0 (i.e. at infinity), we get x+y+z=0 xyz=0 x=0, so exactly the point [0:1:-1:0].
If a=0, from xyz=0, we actually get that this is a triple root, so there is no affine solution.
if w!=0, we dehomogenize w=1 and get to the affine system we wanted to solve. It will now have 2 further solutions (unless a=0), and those are actually easy to find:
Using x=a and rearranging the first equation gives
y=62-a-z
Inserting this into the second equation, we get
a*(62-a-z)z-2880=0
hence
-az^2+a*(62-a)z-2880=0. This is simply a quadratic equation in z. It's roots are real, if the discriminant D=a^2(62-a)^2-4*2880a is nonnegative (in the case D=0, z=y is a double root) and it is rational, if D happens to be a square of a rational number.
Since D is a quartic polynomial with leading coefficient 1, for big and small values of a, the roots will be real.
The only remaining question is, for which rational a this is a rational number.
Writing a as b/c and clearing denominators, we get Dc^4=b^2(62c-b)^2-11520c^3b. The RHS is an integer, the LHS by assumption a square of a rational number. This means that Dc^4 actually already is a perfect integer square. Hence, we're looking for integer squares of the form b^2(62c-b)^2-11520bc with coprime c and b. Note that b is a divisor of the RHS, so it must also be a divisor of the LHS. But then, b^2|LHS, hence b^2|RHS-b^2(62-b)^2=-11520bc^3. Dividing by b yields b|-11520c^3. Since b and c are coprime, b|11520. These are only finitely many cases (divisors of 11520), and I think there aren't too many possibilities for c in each case. Going through these will finally settle the question about rational solutions.
Complex numbers don't fit on the interval [4,40].
Suggested question (I want to learn if this is a known theory in mathematics):
1- If we have an indefinite integral, is there a method to prove that the anti-derivative is expressible/non-expressable in elementary functions?
2- If we have a definite integral, is there a method to prove that it's value is expressible/non-expressable in constants like π, e, √2,... or rational numbers?
Just hit like if you are willing to discuss this question.
As soon as you said that the interval was a hint, I just plugged 4 and 40 into the equation to see if they worked...
They do!
6:12 is it the famous WLOG (without Loss of generality) ?
A further possibility was none are even, but this also leads to a contradiction.
similarly, at around 5:30, shouldn't we consider the possibility that none of x,y,z are congruent to 2 (mod 4)? for instance if two of them are congruent to 1 (mod 4) and the last is a multiple of 4, that would work -- except that it would require the last number to be bigger than 62
@@tylerbreisacher5841 We already proved that all of them have to be even at this point. So 1 (mod 4) no longer applies.
But that possibility has already been proven wrong and it’s trivial.
The product of odd numbers would never give an even answer. I agree that he should have included it but it doesn’t break the world lol
Why not check them all... once you know x is 2 mod 4... eliminate z from the 2 equations to get one quadratic in terms of x and y. Solving for y, the discriminant is x^4-124x^3+3844x^2-11520x = x^2(x^2-124x+3844-11520/x). Possible values of x are 2,6,10,18,30,90. We can eliminate 90 because the sum is 62. Substitute the others into x^2-124x+3844-11520/x and see which are perfect squares. 2 gives a negative discriminant. The other values are (x,discriminant) are (6,1216), (10,1552),(18,1296),(30,640). The only perfect square is 1296.
You can't disregard the bounds for x,y and z, they make the solution unique. Any triple of roots to the cubic x^3 - 62x^2 + Cx - 2880 (C is a free variable) is valid solution, the solution you found is the only one which lies in the interval [4, 40].
@@angelmendez-rivera351 I know I know, I was referring to his comment near the start. He said that he would find all solutions by finding the solutions on [4, 40], which is not the case.
Here are some viewer suggested problems, from my Oxford interview about 20 years ago! (i) if you start at the bottom left corner of an m by n chessboard, how many unique paths are there to the top right corner if you can only move one step right or one step up each go? (ii) how many zeros does 100! have? A lot easier than the Olympiad questions but only had a few minutes to figure them out (but with tutors happy to give hints in places - part of the interview was checking they could actually teach you!)
What if.. we set up the cubic equation t^3 - 62t^2 + mt - 2880 =0, where x, y, and z are its roots (real), and discussed according to the values of the real parameter m their existence and then computed them in that way? Maybe include a little calculus (variations of a the cubic function) in that...
I just wrote up this solution in a different comment.
@@CauchyIntegralFormula nicely done my friend.... i rarely watch at my desk, and so was too lazy to get up and give it a try. Sorry I missed your comment bud.
I don't see where the problem says to take x,y,z to be integers. But if indeed they are, I don't understand why you didn't just do the case analysis. Given x,y,z in [4,40], then by the definitions of a,b,c, we have a in {3,5,7,9,11,13,15,17,19}, b in {1,3,5,7,9}, c in {1,3,5}. Then using the fact that abc = 45, we can further restrict to divisors of 45, giving a in {3,5,9,15}, b in {1,3,5,9}, c in {1,3,5}.
Then if a = 15, bc = 3, so (b,c) is one of (1,3),(3,1), neither of which satisfy b+2c = 8.
If a = 9, bc = 5, so (b,c) is one of (1,5),(5,1). (5,1) doesn't satisfy b+2c=11, but (1,5) does, so we have a solution (a,b,c) = (9,1,5).
If a = 5, bc = 9, so (b,c) is one of (3,3),(9,1), neither of which satisfy b+2c=13.
Finally, if a = 3, bc = 15, so (b,c) is one of (3,5),(5,3), neither of which satisfy b+2c = 14.
So the only solution to a+2b+4c = 31 and abc = 45, with a,b,c odd integers and 2a,4b,8c in [4,40] is (a,b,c) = (9,1,5).
Just so you know, I've made 100's of videos for my math classes, and ending the video has always felt awkward. I'm stealing, "and that's a good place to stop."
He says that at the end of pretty much all his videos.
Problem suggestion, comes from one of the BMO's.
N is a four-digit integer, not ending in zero, and R(N) is the four-digit integer obtained by reversing the digits of N; for example,
R(3275) = 5723.
Determine all such integers N for which R(N) = 4N + 3.
it would be so much better if the problem had a xy+yz+xz=something because that would imply the 3 symmetric sums for vietas formulas on a 3rd degree polynomial.
Hmm, would you not need to prove that x, y, and z have to be integers to say that "x, y, and z must be even"? The interval [4, 40] contains way more than just N \cap [4, 40], after all.
agree
Well if I got it right he kind of assumes that the solutions would be from Integers and eventually shows that the 6 solutions he found are all the possible ones because of a theorem (I honestly don't remember or even know) on polynomial equations...
No. He first only considered only x,y,z integers and at the end argued that this yielded all solutions.
We are just looking for integer solutions first. So You just need to add: if x,y and z are all integers, then they must ...
The final arguments says that this type of problem has 6 solutions max, and as we already we found 6 integer solutions, all solutions are with integers.
Prove that every natural number can be written as the sum of numbers of form 2^m . 3^n where no terms divide each other.
E.g.
23 = 9 + 8 + 6
49 = 27 + 18 + 4
This was the first question on a Putnam exam but I can’t find a video.
without the bounds, there are infinitely many solutions, which you can see by eliminating z and feeding the result into wolfram alpha (which produces a fascinating graph like a three-way parabola with a curved triangle in the middle). the clever part about the bounds is that they cut out a square subset of this graph where a vertical (or horizontal) line will pass through it twice at only a single point - x or y = 18. either way, you necessarily get the other two values at the bounds of 4 and 40. if the solution /didn't/ contain 4 and 40, there would be other solutions in the reals. so with a conspicuous setup like this, a good first guess might be to try the bounds as two of the values and see if that produces a solution.
alternatively: if you fix y = n, you get a quadratic x²n + x(n² - 62n) + 2880 = 0 where the two roots are the values for x and z. but those roots must be within [4, 40], so the question is how to guarantee that. a reasonable guess is to set the midpoint of the parabola at the midpoint of the interval, so (62 - n) / 2 = (4 + 40) / 2 = 22, which gives n = 18 and yields the solutions again. unfortunately i'm not sure how to prove this is the only possibility, since the discriminant ends up being a pretty gnarly quartic.
Thanks - I'm really starting to love the way you do these. Great channel!
Goodness! I just listed the factors of 2880 (4,5,6,78,9,10,...,40) and went through the possible sets from (4,4,180) to (12,15,16). And the only one that summed to 62 was (4,18,40).
I really love your channel sir , I literally need your content
-1
Zerooo
Wow, with these types of problems I'm usually like "well I'll just find one solution (and by symmetry 6 in this case)". No idea I was pretty much done then
Well, you are done finding solutions. But you are not done proving that there are none or no more.
If you wish to show that apart from the ones we have shown, there are none other real solutions all in the interval [4,40], you can do as follows:
1. Show that there doesn't exist a solution in the interval such that x=y (or any other permutation)
2. Define F(x,y,z)=(x+y+z-62,xyz-2880). Suppose there exists a solution (x0,y0,z0) in (4,40)^3, and WLOG x0
What gave you the idea to go mod 4? Lucky guess? Experience? Or was it just one of 100 false starts?
Pretty sure it was just experience, since we had all numbers even numbers, and because 2880 only had 3 prime factors our only real choices were 4,3,9,5.
Maybe the fact that the interval it's [4,40]
No need to get idea that is mod 4 or not
Noted that x,y,z are all even,
Ignore the odd factors
the combinations for x,y,z
are
1.[2a, (2^2)b, (2^3)c]
2. [(2^2)a, (2^2)b, (2^2)c]
3. [(2^4)a, (2)b, (2)c]
the mod 4 test can quickly reject the combination of [2 and 3]
The arguments about x,y,z (mod 4) imply, WLOG, x=4a+2, y=4b, z=4c. The original equations then reduce to: (2a+1)bc = 90 and (2a+1)+2(b+c) = 31.
There are only 10 distinct
factorizations of 90 and it's straightforward to check, for each one, whether (odd factor)+2x(other 2 factors) = 31;
only (9,1,10) qualifies, giving the solution (18,4,40).
What about this problem?
Find all p;q natural numbers such that 1+2*(p^q) is a perfect square
You forgot to mention in the problem that x,y,z are integer....
Here's my solution over the positive integers:
The prime factorization of 2880=2^6×3^2×5.
Because the sum is even, either all summands are even, or one of them is.
But if only one of them is, it must contain all 6 factors of 2, so it must be at least 64, which is too large.
So they are all even so that we can write 2a + 2b + 2c = 62 and 2a2b2c=2880
Or a+b+c=31 and abc=360
Ok, great
Now a,b,c cannot be all odd since abc is even, so one of abc must be odd. There are 2 factors of 2 left, so solutions must have the form
a=e, b=2f, c=4g where e,f,g odd.
Let's see what that gives us.
e+2f+4g=31
efg=45=3^2×5
Fantastic (play with chalk)
g must be smaller than 31/4, so it is either 1, 3 or 5
Case 1. g = 1. We have e+2f=31-4=27, ef=45
1×45 the sum 2+2f is either 47 or 91, both too large
3×15: sum is 21 or 33
5×9: sum is 19 or 23
So case 1 does not give us any solution. Nice. Lets see what case 2 gives us.
Case 2. g = 3. We have e+2f=31-12=19, ef = 45/3=15
1×15: sum is 16 or 31
3×5: sum is 13 or 11
No solution for case 2 either. Great. Finalle move to case 3.
Now let's erase the board (backflip).
Case 3. g = 5. We have e+2f=31-20=11, ef = 45/5=9
1×9: sum is 11. With e=9 and f=1 the e+2f=11 so this is a solution.
Plugging this in we get 2a=2e=18, 2b=4f=4, 2c=8g=40
18+4+40=62 and also 18×4×40=2880. Perfect, this is a solution.
With e=1 f=9 the result of e+2f is 19.
Setting both e and f to 3 gives e+2f=9, fo we have no further solutions and this is a good place to stop.
Hmm I proved that all of x, y, z have to be even. Then for one of them (lets say x) it has to be: 10|x, so x in {10, 20, 30, 40} and then you can check easily for possible solutions.
For example:
X=40
=> y=22-z
=> 22z-z²=72
=> z²-22z+72=0
=> z=11+/- sqrt(11²-72)
=> z=4 v z=18
Very clearer.
Without the interval restriction on values, the solution is the intersection of the two 3-D curves given by the equations in x,y,z. One curve is a plane and the other is a set of hyperbolic shapes. The locus of solutions includes, for example, (3.5, 23.5311402, 34.9688598). Another solution is (-2, -17.64, 81.64) roughly. It would be nice to redo the problem and show the shapes involved and the interval restriction which touches the solution locus at only six points.
The fact that there are only six solutions has nothing to do with number theory and everything to do with the clever construction of the problem.
This is a Floor Function problem from the Indian National Maths Olympiad 2014 problem 2 (since you love floor functions so much)
The problem states : Let n be a natural number , prove that floor(n/1) + floor(n/2) + floor(n/3) + ... + floor(n/n-1) + floor(n/n) + floor(sqrt(n)) is even , below is the link to official pdf of the problem
olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/09/inmo-2014.pdf
Thanks
Just watch this impressive Math channel th-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
@@neelsinha5019 what's RSM?
Didn't we solve this in the comments of a previous video?
@@CauchyIntegralFormula actually i wanted prof michael to make a video on this
X=40 or 18 or 4, Y=18 or 4 or 40, Z=4 or 40 or 18. Easy decision. This is not a hard problem at all. Just plug in 4 and 40 to either xz and solve for y, to xy and solve for z, or to yz and solve for x.
If you disregard the restriction x,y,z є [4;40] there are tons of solutions, if you substitute z for 62-x-y in the second equation you get a third order curve: imgur.com/a/ZoawDle . It is only due to x,y being forced to be in a hexagon by the restrictions 4
From the fact that x,y,z are all even, one may write x=2x', y=2y', z=2z' so that x'+y'+z'=31 and x'y'z'=360. The first equality implies that exactly one, or all, are odd. However, the second equality excludes the second alternative. So one is odd. There is only 3 possibilities in the interval [2,20] : 5, 9, 15. This may help a little bit.
11:50when mom cooks your favourite food
Just for jazz, I did a quick gawk script to check all possible cases on [1,60]. 38 milliseconds later, no other solutions.
What is the best way to suggest problems for you? Do you have an email, I couldn’t figure out how to send direct messages on TH-cam
i dont even know what you did, but even I can see how you did not prove you have all the solutions in those 6 permutations.
6:35 "So if we get a single solution, that's actually 6 solutions…"
Assuming the 3 numbers are distinct. If the numbers were (A, A, B), there are only 3 distinct permutations.
For the record, brute force with a spreadsheet solved this faster than this entire video. Sometimes, brute force is the best way.
I think the brute force only works when assuming integer/natural number solutions - but that is what I did too, but with T-SQL instead: with a as (select * from dbo.fn_sequence(4,40))
select * from a a cross join a b cross join a c
where a.id+b.id+c.id=62 and a.id*b.id*c.id=2880
where fn_sequence is a function that produces a table of integers in the given range.
@@HowardCShawIII Fair, but he stated at 1:00 that we were looking for natural numbers, and I followed his lead.
That's where computer brute force computation shines. Except we need to prove for all of them are natural
Wow, he made this too complex. Recognize that x+y+z=62 is the equation of a spherical shell. So, you have a limited number of solutions. Because x, y, and z are all factors of 2880; x, y and z must be multiples of the base factors or 2, 3, and 5. The interval of 4 to 40 limits your possible multiples for 2 and 3. The power of 3 must be 2 because the multiple must be greater than or equal to 4. The multiple of 2 can be 2, 3, 4, or 5 times. 1 times is less than 4 and 6 times is greater than 40. You can also eliminate 5 as 2880 only has one factor of 5 and that is already accounted for. So, you have to test 2, 3, and 4 powers of 2 for the last number.
I have found all solutions to the system of equations x + y + z = 62 and x * y * z = 2880:
(x, y, z) = (31-z/2+K, 31-z/2-K, z), where K = sqrt[(31-z/2)^2 - (2880/z)] = sqrt[z^3 - 124*z^2 + 3844*z - 11520] / [2*sqrt(z)].
Note that if we solve for K = 0, then we get x = y = 31 - z/2 and z = 1 of the roots of z^3 - 124*z^2 + 3844*z - 11520.
I sometimes get confused by these clips having solutions which seem to be more complicated than they need to be. Within the constraints of 4
I have come to think that the problem was intended over the real numbers, but that Michael just assumed it was over the integers. And that his remark about natural numbers was only intended as a restriction to positive integers.
And it is a much more interesting problem over the real numbers!
For real x,y,z, having xyz=2880 and x+y+z=62 we could pick any value from the given interval for x, and then set yz=2880/x and y+z=62-x.
All solutions then have y^2+z^2 = (y+z)^2- 2yz = (62-x)^2-5760/x.
This corresponds to a circle in the plane with constant x (parallel to the y-z-plane) with radius ranging from sqrt(1940) at x=4 to sqrt(340) at x=40. (The minimum radius occurs at x>62, so well outside the range).
For any choice of x in [4, 40] the circle intersects the square region having y, z in [4, 40] and it should also intersect the line y+z=62-x.
When we pick an x0 with help of Wolfram Alpha, I found the solutions when y
There are going to be other real solutions without the restriction to [4,40]. Your equations reduce to x + y = 62 - z and xy = 2880/z, so x and y are the roots to r^2 + (z - 62)r + 2880/z = 0. The two roots will be real if (z- 62)^2 > 4*2880/z or z(z -62)^2 > 4880. This clearly can happen, so there are definitely other triples of real solutions. You'd need an additional argument to show they can't all be in [4,40].
awesome! @1:00 I paused and factored 2880 to (2^2,2^3*5,whats^left) and they summed to 62........ yea, inside the interval notation.......the FBI calls that a clue as well
Start with: 2880= 2*2*2*2*2*2*3*3*5. The numbers in consideration are: 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 4, 30, 32, 36 and 40. My first choice would be 2880= 4* 720 = 4 * 18 * 40
An Elemantary way..
Since all number are even and there is only one 5...so one of the number must be 10, 20,30 or 40... Forming equations and putting values of z as 10,20,30, 40 we can get values of other numbers
Solving the system:
x+y+z = 62
x*y*z = 2880
gives a one-parameter solution. Taking z as the parameter, and assuming x 40, so these values of z do not give a solution within the required interval. Similarly, for z in (18, 40), x < 4, so these also do not give a solution within the required interval. Only by assigning z a value from the set { 4, 18, 40 } are both x and y also in [ 4, 40 ], so there are indeed only 6 solutions, all of them being some permutation of (x, y, z) = (4, 18, 40).
However, if the restriction to the interval [ 4, 40 ] is lifted, then every value of z such that (z - 62)^2 >= 11520/z leads to real values for x and y, and these (x, y, z) values are all solutions to the system of equations. So, there are, in fact, an uncountably infinite number of solutions, if x, y, and z are allowed to be any real numbers.
no! @Micheal Penn, the interval has a meaning. And why are you considering x,y,z to be naturals?? It would have been a lot easier if you've considered reals. A similar problem on geometry I created, used this problem application. Do have a look at this.
Problem: artofproblemsolving.com/community/c1619209h2357485_geometry_problem__1
Solution: artofproblemsolving.com/community/c1619209h2357490_geometry_problem__1solution
Please use hagaroma chalk for the 100k special vedio
Why does x+y+z = 0mod4 in 5:17?
if x ,y are 2 mod 4 then z has to be 0 mod 4.
At 3.50 why is mod 4 chosen?
I found that this solution was the only one with integers, but I remainded stun by your proof that there are no other real solution : so obvious ... but only when you 've said it !
In fact, he's wrong. There are not just 6 real solutions, the intersection of two surfaces in R^3 is a curve, so there are infinitely many real solutions.
@Angel Mendez-Rivera I mean, his solution/logic is incorrect. There are in fact 6 solutions, but the proof given is wrong. His proof was " Find 6 integer solutions and said that there are only 6 because the max number of real solutions is 2*3=6" That is completely wrong.
Groebner bases solution?
This leads to another problem. If x+y+z = 62 and xyz=2880, then how much is xy + xz + yz?
I got the answer by comparing (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc to x^3 - 62x^2 + Lx - 2880. By setting c = 40, it was a quadratic problem to show a = 4 and b = 18, so L = 952. If L (=ab + ac + bc) equals anything other than 952, then a or c fall outside the bounds. Hard to prove why though.
To clarify: I graphed y = x^3 - 62x^2 + Lx - 2880 in Desmos. After finding that L = 952 by hand, it's clear that any other value of L pushes at least one zero outside the range [4,40]
I do it see above. In fact the unique solution over the naturals is x=4,... And in general. I have x+y+z=62, xyz=2880 then x+y=62-z and x*y=2880/z then 2*x=(62-z)+-√((62-z)^2-4*2880/z). Fix z then there is solution over R iff (62-z)^2-4*2880/z>=0. If z>0 then there is solution iff z*(62-z)^2-4*2880>=0. And if x,y,z>=0 then there is solution iff z*(62-z)^2-4*2880>=0 and 62-z>=0,z
We have 0=f(x)=x^3-62*x^2+L*x-2880. But idea is that L=(2880+62*x^2-x^3)/x and graph L with function of x. We observe with a point a have 3 pre-imagens.
We also have the the discriminant of the polynomial is the form -4*(L^3+a*L^2+b*L+c) then the discriminant is positive over the limited region. If L in Z then the discriminant is a square in the finite points the region. This condition is necessary but not suficient for the roots are integers.
I am a truck driver, so as a matter of professional interest, I think it would be neat if you could derive the equation for a tractrix. It would be a little different from the usual geometry and number theory problems you solve.
That sounds like a nice idea, i'll look into it!
Notice the multiset of prime factors of 2880 is {2,2,2,2,2,2,3,3,5}. (x,y,z) must exhaust this multiset.
(x,y,z) ∈ [4, 40] -> (x,y,z) ∈ {4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 32, 36, 40}
let (a,b,c) be the un-ordered version of (x, y, z)
let a contain exactly one of {5, 10, 15, 20, 30, 40} e.g., it contains the only 5 in the prime factors.
If a=5, then b+c = 57. This means one of the two must be odd and the other even, therefore b=9 and c=38, but 38 is not in the range of possible values, so no (x,y,z) contains 5 as an element.
If a=15, then b+c = 47. This means one of the two must be odd and the other even, therefore b=9 and c=28, but 28 is not in the range of possible values, so no (x,y,z) contains 15 as an element.
If a=10, then b+c = 52. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (16, 36) is the only pair that sum to 52. 10=2*5, 16=2^4, 36=2^2*3^2, the product abc is 2^7 * 3^2 * 5 != 2880 so cannot be a solution.
If a=20 then b+c = 42. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (6, 36) and (18, 24) both pairs contain 3 threes in the join of multiset their multisets so cannot be solutions.
If a=30 then b+c = 32. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (8, 24)) is the only pair that sum to 32. 8=2^3, 24=2^3*3 the produce abc is 2^7*3*5 != 2880 so cannot be a solution.
If a=40 then b+c = 22. Since it is not possible to create two odd numbers from the remaining prime factors, b and c must be even and in the multiset {4, 6, 8, 12, 16, 18, 24, 36}. Notice (4, 18) and (6,16). The product 6*16*40 != 2880 but 4*18*40 = 2880.
This exhausts the possible values for a, therefore all six permutations of 4, 18, 40 are the only solutions to x+y+z=62, xyz=2880, and (x,y,z) in [4, 40]
It's the "unique" answer. But if x,y,z are roots the polynomial f(t)= t^3-62*t^2+r*t-2880. By Vieta's identities then r=(2880+62*u^2-u^3)/u. Where u=x,y or z. Then the function q(u)=(2880+62*u^2-u^3)/u take the value of x there times. Graph q(u) over interval [4,40] and verify the unique r satisfying this is q(4)=q(40).
www.wolframalpha.com/input/?i=%282880%2B62*u%5E2-u%5E3%29%2Fu+from+4++to+40. Graph of q(u). We can prove using methods of fundamental calculus to q(u) but is long.
Good teacher
Please solve this problem
Find the greatest integer less than Or equal to ( sigma n going from 1 to 1599) of 1/√ n
This is from PRMO 2019 right? A lot of similar problems if not exactly the same one asked already on math.stack, you can see there
@@l1mbo69 ok thanks!!
Too many commercials which interrupt the train of thought.
Thanks for all your videos. The way you present the solution is interesting. I have one question for you.
Prove that the number of digits in sum of all n digit natural numbers is 2n.
You understood the question correctly and have presented a good solution
Lalala= 12
Blablabla= 125
Find La and Bla
100k is tangable. It’s coming before the new year for sure
hopefully 2021 is a good year xd
Thank you!
Looking for all integer valued solution would give the same answer. Here is the analysis:
Step 1. Only one is even is impossible since it would imply |x|= 64, |yz|=45, y+z=-33 or 95.
Thus, all are even. Also one of them is divisible by 5.
Hence, without loss of generality, we write x=10a, y=2b, z=2c.
This gives the equivalent system
b+c=31-5a, abc= 8*9
Step2. The case a=y>=z) is
x= 40, y=18, z=4.
@8:25 should be 2^6 on both sides, Michael :)
It went a little above my head.
Didn't understand this 5:45 For what age group is this problem? I'm in high school.
mod is actually quite simply, just watch a few youtube videos. its just the application that is a bit tricky
@@dinkband7662 yes I know mod but it's application is hard. Just need to practice some questions, will get hang of it. It's surprising where it's application pops up.
8:30 I think you meant 2^6 on both sides. Doesn't really matter
Why are we limiting ourselves to integers? The closed interval [4,40] is not the same as the set {4,5,...,40}. And even the notation x,y,z is suggestive of reals as opposed to integers k,m,n for instance.
5*2^a + 9*2^b + 2^c = 62 and a+ b+ c = 6 => a= 3; b = 1; c = 2 so x = 40; y = 18 and z = 4
I was guessing right from the beginning and found the solution quickly. Some of the "2" factors has to be joined with 9 and 5. There are only a few ways you can do this. 5 can be combined with maximum 3 two's (else the sum is higher than 62). 9 can only be combined with 1 or 2 two's (else the sum is higher then 62). So, in the end you've to check only 6 possibilities.
dev way for simple problem -> coding it take 3 lines and 1 min to make
You'd still have to argue why you only consider integer solutions.
@@angelmendez-rivera351 this is the first one where i use a program because i gave up after an hours of reserch
These are one of those questions which u see and think "yaa I can do this" and only to realize later that u have missed many more solutions and u have got some values which don't even satisfy the system(at least happens to me quite often) :p
Wow god is here too
@@BlackDevil-sw9ll yes u R here
Numbers add up to 62 and they range in between 4 and 40.
either two of them are odd or all of them are even
Let x
It is not so easy to factorize 2680 in a snap, buddy...
once you show all are even: let a,b,c = x/2,y/2,z/2.
a+b+c = 62/2 = 31
abc = 2880/8 = 360 = 2^3 x3^2 x5
so the problem has been simplified.
once again we ask about parity for a,b,c.
2 even: even+even+odd = odd is okay, and is the only okay combination.
therefore we know (without loss of generality) a=4d, b=2e.
c+4d+2e = 31
cde = 45
where c,d,e are all odd numbers (in the set 1, 3, 5, 9, 15, 45)
we can eliminate 45 as an option, and begin enumerating.
(1,3,15), (1,5,9), (3,3,5): the only three valid unordered triples.
15+12+2 = 29; 15+4+6 = 25; 3+4+30 = 37; 1+12+30 = 43: no solution with (1,3,15)
1+20+18 = 39; 5+4+18 = 27; 9+20+2=31: one solution with (9,5,1)
3+20+6 = 29; no solution with (3,3,5)
therefore (x,y,z) in any order are (18,40,4)
I want to ask you. when I study from a textbook , should I solve all problems after every section or chapter (too many problems and need much time , also there 's 5 subjects or 6 every semester) or just select some problems which i feel are difficult or have new ideas .
Study until you feel you have mastery. Also, trust your teacher’s judgment on assignments, suggested or required.
@@stephenbeck7222
Thanks for you respond . I meant the selfstudy not the courses i take at university
I just guessed the answer, took me about 10 minutes
Quicker than this lol
Did you prove that you found all?
There are actually infinitely many real solutions. I am not familiar with algebraic geometry so I don't know the theorem he is referring to here but I am pretty sure it does not apply to this situation.
Note that x = 4, y = 18 and z = 40 are in the lattice [4, 40]^3. If x increases, then y and z decrease up until y (or z) reaches 4 at which point you get another permutation on the lattice. So there are infinitely many solutions in the interior of the cube.
I watch the video to see if he found a nice representation of the solutions but ended up being disappointed by all the work he did to reach the solution on the lattice and a wrong conclusion to the problem.
There are negative solutions tho
Am I missing something: A system of two equations within three variables has a one-parameter family of solutions by the implicite function theorem. If these six solutions are the only solutions to the problem, it definitely must have something to do with the interval [4, 40].
You're absolutely not missing anything. This is an intersection of a cubic surface and a plane, i.e. a curve. So there are infinitely many solutions.
@@angelmendez-rivera351 This is true, but any extension of the interval by an arbitrarily small amount would give infinitely many solutions. (Note that all points on the curve are in fact regular)
Maybe you should prove x, y, z are integer first?
Going to touch 100K , today's night...😃😃😃
Bismillaah.
Solve
X+ Y+ Z = 62
XYZ = 2880.
Solution.
First Verify the Factors of
2880. Such that
The Factors Should be three Only and Sum of Factors Should be Equal to = 62.
Then..
The Factors are.
Since 62 is Even.
All the three Factors Should be Even Or Atleast One Factor Should be Even.
2880 is Even.
Here also All the The Three factors Should be Even Or Atleast One Factor should be Even. And Factor should be Multiple of 5 Because Last Digit is 0.(Zero).
Therefore.
One o f three Factors should be Multiple of 5 Or 10.
Some Factors of 2880.
1)80× 4×9.= 2880
But 80+4+9= 93.
2) 40× 8×9 = 2880
&
40 + 8+9 = 57
3) 40× 18× 4 = 2880
&
40 + 18 +4 = 62..
Or
4 + 18+ 40 = 62
It is the the Answer.
Therefore
X = 4. Y = 18 Z = 40
Good