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!
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.
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.
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 :)
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.
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
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
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.
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.
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).
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
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.
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)
@@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.
@@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.
@@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.
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.
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.
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
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" :)
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).
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.
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
@@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.
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
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 :)
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
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.
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.
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?
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?
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.
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
2:17 Good place to start
Antônio Rocha Pereira r u gonna continue this?
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!
Caught me off guard here
That's a good place to answer you comment xd
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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
-math- correction: number theory ...
Fred
Yeah, exactly, just like with the equation n^4 - 598399320112955 n^3 + 5983993201129483 n^2 - 19747177563727221 n + 21542375524066020
= 0
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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.
Nice.
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!
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.
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 :)
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.
So who's gonna make the "Good place to start" account?
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
Not free of rights anymore, it as be done now, but by another person, I guess.
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13:35
Make another channel for Good place to start!🤣
@@TechToppers The set of good places to start is not dense in the set of videos... not yet :p
@@goodplacetostop2973
Okay, after 5-6 videos?
Tech Toppers Not a particular number but once he starts to say it on every video
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
"That's a good place to start" tricked me..
11:35 or make x+y==-z (mod 120) which only leads to 3 cases left to check. might be helpful idk
@Adam Romanov just instead of taking mod 120 take it mod 8 and 5
@@roboto12345 and mod 3
@@roboto12345 yep
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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.
Yup. I did the same method.
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.
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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).
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
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.
that's what I did (after seeing the hints.)
2:16 subverting expectations, flipping the script
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.
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9:27: removing the 1 from the set means the stated equatin is not anymore true.
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)
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
How did you solve it?
@@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.
@@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.
@@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.
Great problem, nice solution!
Thanks Michael.
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.
ua-cam.com/video/Upvk1_M0B-k/v-deo.html
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.
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
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Very reasonable and intelligent solution 😊
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" :)
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1:15
Hey Professor!!
Chalks can be pernicious for your smartboard😁
I know your channel from mathematic world in quora. Remember me as harsh Ranjan that subscribed your channel.
@@rupam6645 Yup 😊
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One solution set is a=4,b=1,c=1,n=4
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).
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.
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
@@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.
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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
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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 :)
I did in this way too.
ua-cam.com/video/Upvk1_M0B-k/v-deo.html
how do you choose the modulo (120 in this case) for these types of problems?
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
ua-cam.com/video/Upvk1_M0B-k/v-deo.html
I didnt understand a freaking thing from ,10:00 can anyone please explain
What is this "mod" in mathematics, can you explain me that with a video here?! ...
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.
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.
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?
hi prof. michael
what is the easiest way to connect with you
i wanted to ask you for a personal advice.
thanks in advance😁
Very nice problem. My solution was identical to yours.
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?
why do you use ℤ (n≥0) instead of using ℕ? is there any reason or just your preference?
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.
We need one account for good place to start and another for Good place to stop
ua-cam.com/video/Upvk1_M0B-k/v-deo.html
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.
ua-cam.com/video/Upvk1_M0B-k/v-deo.html
Modulo again... Can anybody explain what this is? 😭
Remainder
Why don't you teach for math Olympiads, I love your problem solving techniques..... ....
Amazing!
Wow, I came so early that it just uploaded 26 seconds ago
I too tend to come early.
You can make an easier argument using chinese remainders.
Good!!!
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.
Good
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
wow
I was wearing the same shirt today, woot!
Please use disital board parbhuji...,,,🙏🙏🙏🙏
AIME -2
Ttt
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