From now on, Make sure you put your time stamp before good place to stop. That's the second hardest job to do after brain surgery. May the force be with you!
@hamsterdam19424 жыл бұрын
Caught me off guard here
@ccg88034 жыл бұрын
That's a good place to answer you comment xd
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@bowlchamps374 жыл бұрын
Usually math works like this: There are 2 solutions for n=3, 1 solution for n=4. Then there are no more solutions until n=598.399.320.112.945
@ffggddss4 жыл бұрын
-math- correction: number theory ... Fred
@nonamehere96583 жыл бұрын
Yeah, exactly, just like with the equation n^4 - 598399320112955 n^3 + 5983993201129483 n^2 - 19747177563727221 n + 21542375524066020 = 0
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@zanti41324 жыл бұрын
At 10:02, I think a more streamlined approach is possible: Focusing on set B = {1,3,9,27,81}, note that if any number other than 81 is chosen, then even the largest numbers in sets A and C won't get you to 120, i.e. 27 + 64 + 25 = 116 < 120. So, the only chance for another solution has to come when 81 is taken from set B. That means the numbers taken from the other two sets have to total 120 - 81 = 39, and with only three numbers in set C, that possibility is ruled out quickly.
@otakurocklee4 жыл бұрын
Nice.
@emanuellandeholm56574 жыл бұрын
I was thinking of using the set C instead of A (x + y = -z mod 120) instead of A since it's smaller, but your solution is much nicer!
@pyLz4 жыл бұрын
Careful, we want the sum mod 120 to be 0, but 81 = -39 mod 120. Basically, your argument does not take into account that numbers mod 120 have multiple integer representatives.
@zanti41324 жыл бұрын
True, but I figured that was not worth mentioning. All of the integer are positive, so zero and negative numbers are impossible, and if you add the largest numbers in each set you only get to 170, making 120 the only possible multiple of 120 that can be reached. Of course, for a rigorous proof you'd be ruling out even the obviously untrue possibilities :)
@pageboysam3 жыл бұрын
starting at 10:00 max(A) + max(C) = 64 + 25 = 89 -> B >= 31 -> B = {81} -> A + B + C === A + 81 + C === 0 (mod 120) -> A + C === 39 (mod 120) Also, max(A) + max(C) = 89 < 120 -> A + C = 39 (eq. 1) -> A A = {2, 4, 6, 8, 16, 32} max(A) = 32 -> C >= 7, by eq. 1 -> C = {25} -> A + C = A + 25 = 39 -> A = 14 but 14 not in A. Therefore there are no other solutions.
@goodplacetostop29734 жыл бұрын
13:35
@TechToppers4 жыл бұрын
Make another channel for Good place to start!🤣
@goodplacetostop29734 жыл бұрын
@@TechToppers The set of good places to start is not dense in the set of videos... not yet :p
@TechToppers4 жыл бұрын
@@goodplacetostop2973 Okay, after 5-6 videos?
@goodplacetostop29734 жыл бұрын
Tech Toppers Not a particular number but once he starts to say it on every video
@ilyafoskin4 жыл бұрын
I see where this is going... In ten years his video scripts will be entirely composed of specific meme phrases that legacy fans are time stamping, with some math in the background
@monoastro4 жыл бұрын
So who's gonna make the "Good place to start" account?
@goodplacetostop29734 жыл бұрын
The Good place to start meta is not consistent yet. But who knows, the meta has evolved from "That's the final answer" to "That's a good place to stop" :p
@CM63_France4 жыл бұрын
Not free of rights anymore, it as be done now, but by another person, I guess.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@wjx84394 жыл бұрын
11:35 or make x+y==-z (mod 120) which only leads to 3 cases left to check. might be helpful idk
@roboto123454 жыл бұрын
@Adam Romanov just instead of taking mod 120 take it mod 8 and 5
@EliotPlaysMinecraft4 жыл бұрын
@@roboto12345 and mod 3
@pedroafonso83844 жыл бұрын
@@roboto12345 yep
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@ivanlazaro74444 жыл бұрын
In the last step. Instead of checking for all of the elements of C and B for every residue we can argue that, since C and B have only positive numbers, then y+z should be bigger than 120 or that is it possible yo choose y nada z: y+z are exactly equal to every residue (more easy).
@Niglet101bitch4 жыл бұрын
"That's a good place to start" tricked me..
@SSGranor4 жыл бұрын
I found it simpler to consider congruence mod 3, 5, and 8 separately, rather than together, as it allows arguments that depend on little more than the parity of a, b, and c. To wit, 2^n and 5^n are both (-1)^n mod 3; so, the sum will only be divisible by 3 when a and c have opposite parity. On, the other hand, 2^a + 3^b is only congruent to 0 mod 5 when a+b is 2 mod 4 (at worst, this only takes a small amount of casework to check), meaning that a and b must have the same parity. Finally, 3^2 and 5^2 are both 1 mod 8. So, given that the two above conditions combine to require that b and c have opposite parity, 3^b + 5^c must be either 4 or 6 mod 8. The only way for the full sum to be 0 mod 8, then is if a equals 2 or 1, respectively. But, in either case, the parity of a necessary to make the sum 0 mod 8 will be the opposite of the parity of b (i.e. when b is odd, 3^b + 5^c will be 4 mod 8, requiring a=2 and when b is even it will be 6 mod 8, requiring a=1), meaning that simultaneous divisibility by 3, 5, and 8 is impossible.
@Goku_is_my_idol4 жыл бұрын
Yup. I did the same method.
@Goku_is_my_idol4 жыл бұрын
well heres how for the layman to understand: We take n>=5 bcuz it will be ez to put modulo. Next 1. Put mod 2 both sides 1+1=0 (mod2) We dont deduce anything from here 2. Put mod 3 (-1)^a + (-1)^c =0 (mod3) So we gather that if a is even c must be odd and if c is even a must be odd 3. Put mod 4 Since n>=5, 4 divides n! (-1)^b +1=0 (mod4) So b must be odd 4. Put mod 5 2^a +(-2)^b =0 (mod 5) Now this is possible if and only if a=b+2k where k is an integer. Thus since b is odd, a must be odd (from previous steps) Thus c is even 5. Put mod 8 Thus we have a=2k+1 (a is odd) b= 2k+1 (b is odd) c= 2k (c is even) 3^2k=9^k = 1. (Mod 8) 5^2k = 25^k=1. (Mod 8) Now 3^b + 5^c = 4 (mod 8) Adding 2 ^a, where a is odd will never give us 0 (mod 8) which is why there are no solutions for n>=5.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@danielnieuwerf44923 жыл бұрын
Quicker proof for why no solutions for n>4 Take mod 3 to see a and b must have opposite parity. Take mod 5 to see 2^a +3^b can only be 0 mod 5 when a and b have the same parity (4 cases to check). Hence no solutions for n>4
@RexxSchneider2 жыл бұрын
3^0 = 1 and b can take the value 0, so taking mod 3 only shows that a and b have opposite parity when b>0. You have to do more checking.
@louthurston8088Ай бұрын
that's what I did (after seeing the hints.)
@egillandersson17804 жыл бұрын
I worked the same way for the first small solution. After that, I look successively with congruence mod 3, mod 5, mod 4 then mod 8 to show a contradiction. A longer method but less calculations.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@mrflibble57172 жыл бұрын
Great problem, nice solution! Thanks Michael.
@danielrosafranzini92834 жыл бұрын
Congratulations for your work on these videos. You make hard math accessible for anyone willing to do some work to get it. Now, I'd like to suggest a problem for you to make a video: (IMO88-Q6) Let a and b be positive integers such that (ab + 1) divides (a^2 + b^2). Prove that (a^2 + b^2)/(ab + 1) is a perfect square. I have seen a number of solutions (pretty much all of them are hardly disgestable), but it would be great to see one by you, with your great explanations. Thank you.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@c1-math122 ай бұрын
Very reasonable and intelligent solution 😊
@Goku_is_my_idol4 жыл бұрын
Solved it myself by learning from ur earlier videos😊😆😁 I used the same method but i put mod 2,4,8,3,5 all of them individually instead of mod 120 directly(lcm)
@ashimchakraborty29084 жыл бұрын
But it is not because 120 is lcm of 3,8,5. I also thought mod 30 will work well as it is lcm of 2,3,5 but it didn't work
@shreyathebest1004 жыл бұрын
How did you solve it?
@Goku_is_my_idol4 жыл бұрын
@@ashimchakraborty2908 it actually IS bcuz 8,3,5 lcm is 120. Theres a theorem where if a number has some particular given remainders by division from distinct coprime numbers their division from their lcm will give a unique remainder. I forgot name of the theorem but i know for sure that it exists.
@Goku_is_my_idol4 жыл бұрын
@@shreyathebest100 well heres how: We take n>=5 bcuz it will be ez to put modulo. Next 1. Put mod 2 both sides 1+1=0 (mod2) We dont deduce anything from here 2. Put mod 3 (-1)^a + (-1)^c =0 (mod3) So we gather that if a is even c must be odd and if c is even a must be odd 3. Put mod 4 Since n>=5, 4 divides n! (-1)^b +1=0 (mod4) So b must be odd 4. Put mod 5 2^a +(-2)^b =0 (mod 5) Now this is possible if and only if a=b+2k where k is an integer. Thus since b is odd, a must be odd (from previous steps) Thus c is even 5. Put mod 8 Thus we have a=2k+1 (a is odd) b= 2k+1 (b is odd) c= 2k (c is even) 3^2k=9^k = 1. (Mod 8) 5^2k = 25^k=1. (Mod 8) Now 3^b + 5^c = 4 (mod 8) Adding 2 ^a, where a is odd will never give us 0 (mod 8) which is why there are no solutions for n>=5.
@nicolasnauli86584 жыл бұрын
@@Goku_is_my_idol I don't think you can assume 2^a is always divisible by 4 when doing mod 4. It could be 1 or 2. Though 1 could be ruled out since that would make n! odd. As for the case if 2^a is 2, you need to do the same working (with the exception that b is even) and also for mod 5, 5^c can also be 1 which further complicates things. Oh, I guess this also applies to mod 3 actually. Hmm.
@keinKlarname4 жыл бұрын
9:27: removing the 1 from the set means the stated equatin is not anymore true.
@leickrobinson51864 жыл бұрын
BTW, an easy way to narrow down the cases that you need to check at the end is to note that (for n>=4) taking mod 4 of the equation immediately leads to either {a=1 and b is even} or {a>1 and b is odd}. :-D
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@CM63_France4 жыл бұрын
Hi, I don't know if it is the first time or not, but in his last video, Mathologer ended by saying "that's a good place to stop". As we say in french : "un clin d’œil" :)
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@MathrillSohamJoshi4 жыл бұрын
1:15 Hey Professor!! Chalks can be pernicious for your smartboard😁
@rupam66454 жыл бұрын
I know your channel from mathematic world in quora. Remember me as harsh Ranjan that subscribed your channel.
@MathrillSohamJoshi4 жыл бұрын
@@rupam6645 Yup 😊
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80😊
@binamahadani32673 жыл бұрын
One solution set is a=4,b=1,c=1,n=4
@composerlmythomorphic26354 жыл бұрын
For n>=5, Notice n!==0 (mod 4) and 2^a+3^b+5^c==2^a+(-1)^b+1==0 (mod 4). This is possible only if [I]: {even a, odd b} OR [II]:{odd a, even b}. Also notice n!==0 (mod 5) and 2^a+3^b+5^c==2^a+(-2)^b==0 (mod 5). Substituting case [I] and [II] into the mod 5 equation and no possible solution is found.
@jesusthroughmary4 жыл бұрын
2:16 subverting expectations, flipping the script
@laszloliptak6114 жыл бұрын
Small nitpick: While you can ignore the possible remainder 1 in set A when checking if there are solutions, it is still an element of A, so you can't cross it out.
@ffggddss4 жыл бұрын
It's being crossed out as not leading to a solution, because it would make the desired equation impossible to satisfy; he isn't "expelling" it from set A. That's totally legit. Fred
@laszloliptak6114 жыл бұрын
@@ffggddss I understand that he did not mean to remove it from the set, that is why I called it a nitpick. Crossing out something in a solution means either that it cancels with another term or that it was written in error and we don't want it to be included.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@riadsouissi4 жыл бұрын
I think mod 120 is a bit brute force but works great here. I went differently. Beside identifying the two first obvious solutions, I started with mode 2, mode 3, mode 5, tried other few mods but nothing helped. Then tried mode 4 (2^2), and here a nice restriction on the power of 3 appeared but not good enough. Then it was a clear hint to try mode 8 (2^3) and here both powers of 3 and 5 were constrained. After that, back to mode 3 to add another constraints on the power of 2. Ended up with this equation: 16*4^a+3*9^b+5*25^c=n! for n higher than 3. And finally mode 5 to show all combinations lead to no solutions except when the constrained powers a, b, c of the new equations are = 0 which leads to the last solution (4,1,1,4).
@mathsaddict70294 жыл бұрын
hi prof. michael what is the easiest way to connect with you i wanted to ask you for a personal advice. thanks in advance😁
@surem83194 жыл бұрын
I solved it a different, but similar way (also with using modular algebra). I found rules every step of the way until a "conflict" would occur. For a number on the form n! all numbers from 1 up to n must divide it, so we can build some rules. The lowest number we can get (a,b,c)=(0,0,0)=3 is bigger than 2! so there is no solution for n0 and must both be even for b = 0. Next up 4!=24. Here we find the solution (4,1,1). From mod 4 we get rule: 'a' and 'b' must have opposite parity. (which also means that 'b' and 'c' must have the same parity including b = 0 from the rule for mod 3 since 0 also is even) Now we get to mod 5. Let's first assume that c>0 then: 2^a + 3^b + 5^c (mod 5) congr to 2^a+3^b mod 5 congr to 0. Looking at the residues we get 2^a mod 5 cong to {1,2,4,3,1,2,4...} and 3^b mod 5 cong to {1,3,4,2,1,...} the only way these can sum to 5 is by having the combinations 2+3 or 4+1, but both of these end up having a value for 'a' and 'b' with the same parity which we can't have (due to the rule from mod 3). Therefore, if there is a solution, c = 0: 2^a + 3^b + 1 (mod 5) cong to 0. From this we now want 2^a+3^b cong to 4 mod 5 which can only be done by 2+2 or 3+1. 2+2 Has the same parity and can be quickly ruled out and for 3+1 we must have the same parity for 'b' and 'c' (from the rule of mod 4) so b=1 is a necessity while a=3. This does not have a solution though. Since we can't get a number that is divisible by both 5 and the numbers before it at the same time we got no hope for any future numbers either. Therefore the three we found are the only possible ones. Probably not the most beautiful solution, but it was a fun puzzle nonetheless :)
@nilavadebnath24254 жыл бұрын
I did in this way too.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@tomatrix75253 жыл бұрын
At the final modular equivalence, would we need to work all them out for full marks? To me it’s clear that no sum of x, y,z is a multiple of 120(i.e. 0 mod 120) but of course that’s not like rigid. Could you just write that no x,y,z satisfies it?
@youknowme33462 жыл бұрын
I didnt understand a freaking thing from ,10:00 can anyone please explain
@a_llama4 жыл бұрын
how do you choose the modulo (120 in this case) for these types of problems?
@hiddeneagle14084 жыл бұрын
In general, it is intuition. In this case, if you play around with n bigger than 4, you wouldn't get any other solutions. So in this case we want to prove that there are no solutions for n bigger or equal to 5. Since n! is what the left side of the equation should be equal to, comparing it to n! would give us the best way to do it. That's the way I understand it
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@ffggddss4 жыл бұрын
First off, a, b, and c all have to be ≥ 0 for the LHS to be an integer; n ≥ 0 for the RHS to be finite. This makes the LHS ≥ 3, which will require n to be ≥ 3. Quickly found then, for n = 3, n! = 6, are: (a,b,c,n) = (1,1,0,3); 2+3+1 = 6 and (2,0,0,3); 4+1+1 = 6 and no other solutions with n=3. For n = 4, there's (a,b,c,n) = (4,1,1,4); 16+3+5 = 24 All other possibilities for n=4 can soon be exhausted, with no solution found. Now with n ≥ 5, n! will always be divisible by 5! = 120, and therefore, by 30. But powers of 5 with exponent c ≥ 1 are all ≡ ±5 mod 30. There may be some modular base that will help here; I haven't found it yet. I suspect there are no other solutions, but can't yet show that. Let's see what MP has up his sleeve... Ah! So he went all the way up to mod 120, and did a proof by exhaustion. Bravo! Fred
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@youknowme33462 жыл бұрын
I didnt get it if there are solutions to the given equation where n≥5 then why does it also give solution to x+y+z=0(mod 120) i mean x,y,z are just the residues of the powers mod 120 also y+z=-x how the does y+z=118 i mean how the freakin hell on earth?
@tomatrix75253 жыл бұрын
Very nice problem. My solution was identical to yours.
@lexyeevee4 жыл бұрын
it's not hard to convince oneself there are no workable triplets without manually checking them all: the biggest elements in B and C are 81 and 25 which add to 106, so sums of 112, 116, and 118 are impossible; 104 is close, but the next-biggest elements in B and C are much smaller so there's not enough wiggle room to make that; with the 81, you'd need 7 to make 88, which you don't have, and without the 81 the best you can do is 52, so 88 is out; and for much the same reason, you can't get any closer to 56 than 52.
@samu8044 жыл бұрын
Amazing!
@stefan31983 жыл бұрын
why do you use ℤ (n≥0) instead of using ℕ? is there any reason or just your preference?
@RexxSchneider2 жыл бұрын
It's because he uses the convention that 0 is not an element of ℕ. So he has to state explicitly ℤ_n≥0 for the set of non-negative integers.
@mrrashedali4 жыл бұрын
We need one account for good place to start and another for Good place to stop
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@klementhajrullaj12223 жыл бұрын
What is this "mod" in mathematics, can you explain me that with a video here?! ...
@RexxSchneider2 жыл бұрын
Start with the overview at en.wikipedia.org/wiki/Modular_arithmetic and follow any of the links in the Notes, References, and External links sections for more info.
@garvittiwari11a614 жыл бұрын
Why don't you teach for math Olympiads, I love your problem solving techniques..... ....
@robertgerbicz4 жыл бұрын
Another proof for n>=5. Suppose that a>=2,b>=1,c>=1: 2^a+(-2)^b==0 mod 5 2^a+2^c==0 mod 3 (-1)^b+1==0 mod 4 From these b is odd, a is odd, c is even from this a>=3 so 2^a+3^b+5^c==0+3+1==4 mod 8 contradiction. We left: a0) 2+3^b+5^c=n! b is even (use mod 4) c is even (use mod 3) 2^a+3^b+5^c==2+1+1==4 mod 8 contradiction. With this the proof is complete.
@chandranichaki95802 жыл бұрын
kzbin.info/www/bejne/i6HZnGSVgpV1Y80
@kartikdas86113 жыл бұрын
Good!!!
@demenion35214 жыл бұрын
that's a really clever trick to reduce mod 5! as then obviously the powers of 2, 3 and 5 only form small subsets of all residues.
@user-A1684 жыл бұрын
Good
@СВЭП-и4ф11 ай бұрын
mod 3 = 2^a + (-2)^c, c must be odd, mod 4 = (-1)^b + 1^c => b and c are opposite parity, mod 5 = 2^a + (-2)^b, b must odd, so n! can’t be divisive by 5 (for a > 1). For a=1 b must be even, 3^2k= 9^k = 4 or 1 mod 5, 2 is 2 mod 5, 5 is 0 mod 5, no solution
@redshibi4 жыл бұрын
You can make an easier argument using chinese remainders.
@linggamusroji2274 жыл бұрын
Wow, I came so early that it just uploaded 26 seconds ago
@TikeMyson694 жыл бұрын
I too tend to come early.
@mcbeaulieu4 жыл бұрын
Modulo again... Can anybody explain what this is? 😭