Chile Mathematical Olympiad | 2011
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3 жыл бұрынWe give a solution to a nice number theory problem from the 2011 Chile Math Olympiad.
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@fountainovaphilosopher8112 3 жыл бұрын
There is a small mistake here. If a is 3, then 1/b+1/c = 3/4- 1/3, which is 5/12, not 7/12. Because of this, another valid solution ( (a, b, c)=(3, 4, 6) ) is missed out. Interesting problem however.
@sahilbaori9052 3 жыл бұрын
Exactly.
@SREproducciones 3 жыл бұрын
I think that (3, 3, 12) is another solution that falls from that case.
@haraldsbaumanis 3 жыл бұрын
@@SREproducciones This solution is already in the video. Indeed the only one he missed because of the mistake was (3, 4, 6)
@gabonafold9446 3 жыл бұрын
In the case a=3, let b be be equal or smaller than 2. From that you can se that 1/b +1/c is bigger or equal to 1/2 +1/c>1/2>1/2.4=5/12. by contradiction that means that b>2.So, if b=3, then 1/3+1/c=5/12, and from that you can conclude that c=12. So, when a=3, there isn't just the solution (3;4;6), but the (3;3;12) solution too. In total, I got the solutions:(2;5;20),(2;6;12),(2;8;8),(3;3;12),(3;4;6),(4;4;4;). Correct me, if I'm wrong.
@nicholasfrieler5005 3 жыл бұрын
@@gabonafold9446 One small mistake (it probably was just a typo), (2,4,20) is not a valid solution but (2,5,20) is.
@mohamedaminehadji6415 3 жыл бұрын
There is a small mistake at the end. If a = 2 and b = 7, then c = 28/3 which is not a natural number (Because 28/7 is a natural number)
@swerasnym 3 жыл бұрын
Lots of small mistakes in this one: 1) 3/4 - 1/3 = 9/12 - 4/12 = 5/12, so in case 2 no solutions if b >= 5 (not 4) and giving (a,b,c)= (3,4,6) as a missed solution. 2) In the last case (a,b) = (2,7) we get 1/c = 1/4 -1/7= 7/28 - 4/28 = 3/28 i.e. c= 28/3 which is not in N. Solution: (a,b,c) \in {(4,4,4), (3,4,6), (3,3,12), (2,8,8), (2,6,12), (2,5,20)}
@sanchayitamukherjee7906 3 жыл бұрын
And the mistake misleads further course of action
@egillandersson1780 3 жыл бұрын
A little mistake : 3/4 - 1/3 = 5/12 (and not 7/12) So, there is another solution (3, 4, 6)
@leickrobinson5186 3 жыл бұрын
Good! That’s what I got, too. :-D
@NightRider0101 3 жыл бұрын
I was wondering if I forgot subtraction of fractions.
@Walczyk 3 жыл бұрын
yeah lol i just was stuck here for a minute
@Walczyk 3 жыл бұрын
wait i got 1/3 + 1/3 + 1/12
@damonpalovaara4211 3 жыл бұрын
3, 3, 12 also works but that's it for the A=3 case
@julianosiniri 3 жыл бұрын
Why isn't the solution so simple as (4, 4, 4)? 1/4 + 1/4 + 1/4 = 3/4
@JB-ym4up 3 жыл бұрын
All solutions, that is 1 of 6.
@chandy3859 3 жыл бұрын
4:00 and the question did say "find all"
@martinator9480 3 жыл бұрын
Also adding onto the comments you can't make a=b=c afaik
@philos22 3 жыл бұрын
@@martinator9480 why not?
@chandy3859 3 жыл бұрын
@@martinator9480 look at 4:00
@williamadams137 3 жыл бұрын
8:36 Why is 28/7 not a natural number? 🤷🏻♂️
@williamadams137 3 жыл бұрын
4:44 Why is 3/4 - 1/3 = 7/12 ? 🤷🏻♂️
@minh9545 3 жыл бұрын
Mistakes happen sometimes, hopes he see this comment.
@alexblm9943 3 жыл бұрын
That's 28/3, which is not a natural number but yeah that's confusing
@reynaldopanji2066 3 жыл бұрын
28/7 = 4 Natural number, but not if it was 28/3
@IoT_ 3 жыл бұрын
@@reynaldopanji2066 it is actually 28/3 , not it WAS. Probably it was a typo 😅
@noelani976 3 жыл бұрын
A forgivable maths computation error I want to point out here, and that was towards the end of this great video. The case when a=2, b=7, the value for "c" would be 28/3 and not 28/7, which will give you the figure "4"
@andreamarino95 3 жыл бұрын
There's a small argument one can do to avoid cases once you have chosen a=2, 3,4. Indeed, when you are left with an equation like 1/b+1/c = p/q, then multiplying by bcq you get cq+bq = pbc c= bq/(pb-q) This kind of divisibility is easier to solve, because you can take out a multiple of the denominator to get rid of the part depending on the variable. In fornulas: pc = pbq/(pb-q) -q +q = q^2/(pb-q) Now for every divisor d of q^2 check whether pb-q = d has an integral solution b, and you are done. For example, a=2 case yield p=1, q=4. b-4 = divisor of 16 has solutions b=2, 3,5,6,8,12,18
@vittoriolufrano9814 3 жыл бұрын
You are awesome! Hope that you will do more geometry /combinatory problem such that the IMO one, thanxs
@johnpoole6711 3 жыл бұрын
Bad math. Makes elementary mistakes.
@juanmirpieras 3 жыл бұрын
3/4-1/3=5/12, and 28/7=4
@MORISENSEIISGOD 3 жыл бұрын
Hey Michael, loving these videos. Keep it up! I found some interesting problems from past AIMEs. They are 2015 AIME I problem 13, and 2015 AIME II problem 13, both to do with summing, or multiplying trig functions.
@alexbrodbelt297 3 жыл бұрын
Dude, this channel is amazing
@minh9545 3 жыл бұрын
How about trying the second and third problem in IMO 2020?
@johnloony68 3 жыл бұрын
I worked it out by visualising a clock face. To get three fractions adding up to 3/4, it's like three slices of a circle adding up to get to the 9 o'clock position (or, if you prefer, 45 minutes). The smallest possible size for the largest slice is 15 minutes, so 1/4 + 1/4 + 1/4 = 3/4. The second solution is 4 o'clock, 8 o'clock and 9 o'clock (3,3,12). The third solution (where he made a mistake) is 1/3 + 1/4 + 1/6. Or 4 o'clock, 7 o' clock and 9 o'clock... or 20 minutes, 15 minutes and 10 minutes.
@akhotaba7866 3 жыл бұрын
Try this problem find all solutions for the following equation: 1/a + 1/b + 1/c + 1/d = 1 where a,b,c,d are integers > 2
@akhotaba7866 3 жыл бұрын
@@angelmendez-rivera351 you had a little mistake where u got d = 10 but its actually 12 . then, the solutions of the form (a,b,c,d) are: (3,3,4,12) (3,3,6,6) (3,4,4,6) (4,4,4,4) up to permutations.
@leadnitrate2194 3 жыл бұрын
@Cactus Jack yeah, this is big brain time
@caesar_cipher 3 жыл бұрын
actually here there are no solutions for integers 1 or 2 - so u will get the same solutions without your condition
@rogierbrussee3460 3 жыл бұрын
Ah, you may not realise, but this little problem contains in it the ultimate question, to the answer of life and the universe. the very reason this planet exists, at least according to nerdy friends who keep coming up with the largest possible value for d (assuming a
@noelani976 3 жыл бұрын
@@akhotaba7866 31 solutions in total. That's good!
@NicolasCageIsActuallyARobot 3 жыл бұрын
I'd love to see this solved using dual simplex or constraint linear programming as well! This is a great intro to Operational Research problem.
@unnamedtoaster 3 жыл бұрын
I wonder if the has applications in electrical engineering or physics. If a, b, and c were resistor values, then the total resistance if you put resistors a, b, and c in parallel would be 4/3.
@demenion3521 3 жыл бұрын
even though the solution is very clear, i think it would've been better to derive the bounds on a and b without just stating and proving them. for example 3/4 = 1/a+1/b+1/c a
@wasfiiwasfi 3 жыл бұрын
i love all your videos !
@thayanithirk1784 3 жыл бұрын
Please try some China team selection test problems it is said to be tougher than IMO
@ezioarno15 3 жыл бұрын
So the solutions are (2. 5, 20) (2, 6, 12) (2, 8, 8) (3, 3, 12) (3, 4, 6) (4, 4, 4)
@manuelg758 3 жыл бұрын
Correct.
@gmchess7367 3 жыл бұрын
The last question that you have given there are coming more than 30 cases or many more so how to do it easily please help anyone ????
@marshalls36 3 жыл бұрын
Eq.1) a+b/ab = 3c -4/4c Similarly, Eq.2) b+c/ab = 3b-4/ab Eq. 1- Eq. 2 => c-b/bc = a-b/ab Get a=b =c = 4 Consider a = c, Substitue into Eq. 1/a + 1/a + 1/b, get 3a-8 factor under, then try a = 3, get a = c =3, b =12
@HagenvonEitzen 3 жыл бұрын
To combine a lot: If a >= 4 then 1/a + 1/b + 1/c >= 3/a >= 3/4 with equality iff a=b=c=4
@ramaprasadghosh717 3 жыл бұрын
Mathematics is the logic of elegance . There is no point of going overkill for anything
@caesar_cipher 3 жыл бұрын
This video must get a record for comments pointing the same 2 mistakes over and over and over .... I don't understand these commenters' psychology - they see that the errors have been posted by previous users and yet have the deep-seated urge to repeat ??!! Wtf is wrong y'all ?
@egillandersson1780 3 жыл бұрын
@@angelmendez-rivera351 No arrogance at all : just not seen the previous comment before writing mine. 😕
@ribozyme2899 3 жыл бұрын
Someone else already commented on this issue before you. Why did you feel the need to write this anyway?
@ribozyme2899 3 жыл бұрын
@@angelmendez-rivera351 Why don't you tell that to those whom you complained about instead of those who already agree with you?
@caesar_cipher 3 жыл бұрын
@@ribozyme2899 After reading your comment, I went back and read all the repetitive blabber sorted by time. I didn't see anyone before me highlighting the needlessness of repeating the same corrections over and over again. So your comment is misinformed.
@caesar_cipher 3 жыл бұрын
@@angelmendez-rivera351 agree mate ! By all means its great to see alternative solutions, even if they are not the easiest or even "overkill". Also extensions to a problem or new problems also enrich comments. Even smart memes like "good place to stop" or math jokes are always welcome. But repeating the same errata point is like such nuts & irritating ! That's why stack exchange is such a great forum where without a certain minimum reputation points one cannot comment, keeps away self-righteous users from repeating junk
@LorxusIsAFox 3 жыл бұрын
I think you've made a mistake in the a = 3 case - this means you end up missing the solution (3, 4, 6), and need to do a little more work on the subcases for b. Also, in the a = 2, b = 7 subcase, you should be getting that it's (edit) 28/3 that isn't an integer.
@MarcoMate87 3 жыл бұрын
No, c = 28/3.
@LorxusIsAFox 3 жыл бұрын
@@MarcoMate87 Ah, you're right.
@megauser8512 3 жыл бұрын
Yep, I get this too!
@georgesbv1 3 жыл бұрын
For the homework: 1/a+1/b+2/c+3/d=5/7 By limits we can establish that 2
@pabloAT98 3 жыл бұрын
Nice solution, cheers from Chile!!
@annog6673 3 жыл бұрын
Could you please solve problems like this without the solution? That would show us your approach to those problems. Would love it!
@xxxy7232 3 жыл бұрын
Did anyone find a solution for the last question?
@bloodhunter4487 3 жыл бұрын
When a=2 and b=7 then c=28/3 not 28/7 And 28/7 is a natural number and equal 4 .
@juanpablogonzalez7130 3 жыл бұрын
Yahoo!! I'm Chilean. Viva Chile =)
3 жыл бұрын
From Barranco lima Peru, my son give me this chanel that is very nice and good exercices to share
@savagezerox 3 жыл бұрын
3/4 - 1/3 doesnt equal 7/12. it equals 5/12. @4:35 3/4 * (3/3) - 1/3 * (4/4) = 9/12 - 4/12 = 5/12, not 7/12. you then say no solutions if a=3 and b>=4, but a=3, b=4, and c=6 is a solution.
@anshumanagrawal346 3 жыл бұрын
Ya, I also noticed that he missed a solution there because of that mistake
@coreybailey9362 3 жыл бұрын
5:31 Since 3/4 - 1/3 = 5/12 and not 7/12, this means that b=4 is a possibility. You can do the same argument and find out that b has no solution when b is greater than or equal to 5. Excluding b=4 when a=3 causes you to miss a solution to this problem. The solution where (a,b,c)=(3,4,6).
@user-ue5gl5hh4c 3 жыл бұрын
Amazing! Thank you very much. Can you make a full solution of a Mathematical Olympiad test? It would be awesome. Thank you very much
@goodplacetostop2973 3 жыл бұрын
9:01 Es un buen momento para parar
@goodplacetostart9099 3 жыл бұрын
buen lugar para empezar 0:00 👍
@zeravam 3 жыл бұрын
¿En serio?
@IoT_ 3 жыл бұрын
Please write down in Russian next time: "И это хорошее место, чтобы остановиться" 😁
@goodplacetostop2973 3 жыл бұрын
Eg. M Sure, next time Michael shows a math challenge from Russia 😁
@danielrc9016 3 жыл бұрын
Plsssss start writing in the language of the olympiads plss
@zooboor8148 3 жыл бұрын
2:27 it couldn't be greater or equel to 1 because a,b,c is in the natural number and there's no natural number that presents b or c ant exsits this equation
@anshumanagrawal346 3 жыл бұрын
What?
@aniketsen6845 3 жыл бұрын
Sir, i was able to solve this problem in a more simple way..and the problem was simple and interesting! Your channel is a very good source of good questions in mathematics....Loved your hardwork sir🙂😀 🖤LOVE FROM INDIA🖤
@TechToppers 3 жыл бұрын
Your way?
@stvp68 3 жыл бұрын
Can you use QED for these solutions, or is that limited to a different type of problem?
@jomama3465 3 жыл бұрын
It's mostly used for formal proofs and even in such formal proofs, it's rarely used.
@stvp68 3 жыл бұрын
Ron Matthew Austria That makes me sad. I college math courses, that was always our favorite thing: to be able to say QED after successfully finishing a proof. Happy memories!
@ribozyme2899 3 жыл бұрын
QED means "qod errat demonstrandum", "as was to be shown". I.e., you use it after proving something, (usually) not after finding previously-unknown solutions.
@TechToppers 3 жыл бұрын
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper
@flobiish 3 жыл бұрын
The only way to get 4 in the denominator being abc are natural is if abc are factors or multiples of 4, i.e. 2 to a power. In some order we need 1/2, 1/8 and 1/8. Easily done by binary calculations.
@melkenhoning158 3 жыл бұрын
a = 2 b = 6 c = 12 -> satisfies 1 ≤ a ≤ b ≤ c. 1. Is this correct? 2. It says find ALL, so does that mean my answer is just one of many?
@damjantarkanyi9479 3 жыл бұрын
Yes. Yes.
@melkenhoning158 3 жыл бұрын
@@damjantarkanyi9479 Ah thanks!
@Em-bb1mn 3 жыл бұрын
Hi. A couple errors in this video but that can be corrected easily. If b=7, you say c = 28/7 but 28/7 = 4 so it works. It’s just the value of c which is erroneous. And in the bounds of your second case you bound it by 7/12 when it should be 5/12 I think
@admiralcasperr 3 жыл бұрын
How are one exactly is supposed to guess the a>=5? This is a bad Olympiad problem since it's not hard but just a gamble.
@alasanof 3 жыл бұрын
Before watching the video, the solutions I found were 1/2+1/6+1/12; 1/3+1/4+1/6. I found them by splitting 3/4 into 2/4+1/4, which is 1/2+1/4. Then you can get an equivalent fraction by multiplying one of them by 3, we can get 1/2+3/12, to 1/2+2/12+1/12, to 1/2+1/6+1/12. I didn't see the signs were ≤ instead of
@duanedonaldson2262 3 жыл бұрын
1/4 + 1/3 + 1/6 = 3/4 which is the equal equation of 3/12 + 4/12 + 2/12 = 9/12 = 3/4 SO a=3, b=4, c=6 which fits into a similar CASE 2. Finally an easy problem on this GREAT Channel. 1/6 + 1/42 + 2/21 + 3/7 = 5/7, the first 2 terms equal 8/42 as my theme was to keep all multiples of 7 and do not reuse any number twice, when removing 1/42 then 7/42 was the remaining which simplified to 1/6, best I could do.
@TechToppers 3 жыл бұрын
You didn't show that they are unique. How sure you are that they are unique?
@ramanakv3272 3 жыл бұрын
1/a,1/b,1/c are in increasing order 3/4 lies between 3/a,3/c which gives a=c=4 hence b=4
@trailmixvideo 3 жыл бұрын
I found the solution 3,4,6 while I was going to sleep last night. Then I remembered it today and thought I would check the video to see the solution. I was disconcerted when he demonstrated that b could not be 4. I had solved the problem using a common denominator of 24 after making some estimates of the bounds on the solution. I was sure that the solution would offer some insight into using the Chinese Remainder Theorem, or would incorporate some methods of approaching word problems. After learning that my solution, that is clearly true when you convert to the lcd of 24, was not to be considered, I was so taken aback that I paused the video and went directly to the comments. Apparently the methodology led to a detour around the solution.
@rohithiitgn 3 жыл бұрын
4:42 It should be 5/12 right?? Not 7/12
@ramk4004 3 жыл бұрын
Please do more combinatorics problems sir
@1114maniacmike 3 жыл бұрын
I was lost when he brought out the 5. Where did that come from?
@JohnDoe-mj6cc 3 жыл бұрын
It was a random guess, but one he had prepared beforehand. The five is important because it gives some limits
@hybmnzz2658 3 жыл бұрын
Intuitively you would guess the left hand side becomes to small when a gets too large. The specific value of 5 may require some experimentation.
@anshumanagrawal346 3 жыл бұрын
Guess n check
@diddierhilarion 3 жыл бұрын
4:56 isn't it 5/12 ?
@bibeksubedii0001 3 жыл бұрын
i mistake : 3/4 - 1/3 = 5/12 (and not 7/12) So, there is another solution (3, 4, 6) In the last case (a,b) = (2,7) we get 1/c = 1/4 -1/7= 7/28 - 4/28 = 3/28 i.e. c= 28/3 which is not in N. Solution: (a,b,c) \in {(4,4,4), (3,4,6), (3,3,12), (2,8,8), (2,6,12), (2,5,20)}
@ayoubslaoui7530 3 жыл бұрын
{a,b,c} = {3,4,6} is missing. Other than that. Great channel. Keep up the good work !!
@drx_ander 3 жыл бұрын
There are no rules for the solution at the a-b-c-d-problem been made... So here is the first one: a=2, b=5, c=141, d=29610 I think there will be many more...
@holylampposts 3 жыл бұрын
You guys ever notice that mathematicians say "but" when they mean to say "and?" I wonder why that is. I noticed it just started to happen with my classmates at some point when I was in undergrad. I couldn't tell what prompted it.
@MrRyanroberson1 3 жыл бұрын
mathematicians say but when it contradicts a statement that precedes it the same way everyone does.
@holylampposts 3 жыл бұрын
@@MrRyanroberson1 Agreed. Mathematicians also say but when a statement that follows a previous statement is a noncontradictory result of the previous statement. Consider the prof's statement, "If a >= 5, then we have b and c are also >= 5, but then, taking the reciprocal..." Why but? The following statements are noncontradictory to the preceding statements. No need for but, just say, "and." Consider also, "if a = 3, then that's going to make 1/b + 1/c = 3/4 - 1/3, but that's going to be equal to 7/12." Again, why but? There's no contradiction. Just a direct result of the preceding information.
@MrRyanroberson1 3 жыл бұрын
because "but then DOING x" is meant to signify that the following action is an attempt at causing a contradiction (example: if a > 5 then b, c > 5, but then 1/a + 1/b + 1/c < 3/5 < 3/4, which is a contradiction). though i did not notice the second example for 7/12 which does appear to be a false use of "but"
@siddheshsingabhatti4025 3 жыл бұрын
Either (4,4,4) or (3,4,6) maybe.. Am I right ??
@flobiish 3 жыл бұрын
Presuming a unique solution, 1/a + 1/b + 1/c = 3/4: the simple solution would be: a=2, b and c = 8. 1/2 + 1/8 + 1/8 = 3/4. Adding any prime factors to the denominators would complicate things.
@gajjarsachin4795 3 жыл бұрын
3,4,6 is also a solution, which is not mentioned.
@ffggddss 3 жыл бұрын
Looks like you've got a couple or so corrections to make here, Michael... Fred
@its_shxblvrbs1786 3 жыл бұрын
Am I crazy, or can you just make a, b, and c all equal 4 and call it a day? (Edit): I guessed that before the video started for me😂
@Mischkovonik 3 жыл бұрын
Why not a = b = c = 4 ?
@moezbahri6380 3 жыл бұрын
In case 3,c=28/7 is integer but not bigger than b=7
@denis3141 3 жыл бұрын
Me resulta mas fácil pensarlo como números decimales ( ? )+( ? )+( ? ) = 0.75 0.25+0.25+0.25 = 0.75 lo que es igual 1/4+1/4+1/4 =3/4
@JuanEsquivel-ex8nv 3 жыл бұрын
Claro, el problema es conseguir los otros casos.
@xwtek3505 3 жыл бұрын
I initially used a cubic formula, only to encounter overly complex formula (basically I used vieta formula)
@fahad-xi-a8260 3 жыл бұрын
8:36 It should be c = 28/3 not 28/7 which is clearly a natural number 4. Happens. I too can not sometimes understand my own hand writing.
@bradhoward 3 жыл бұрын
28/7??
@mach2570 3 жыл бұрын
When a = 3, 1/b + 1/c = 5/12 NOT 7/12. For a = 3 we get solutions (3,3,12), (3,4,6) .
@ameerunbegum7525 3 жыл бұрын
Amazing.
@carpyook 3 жыл бұрын
Could anyone confirm that (1/a+1/b+2/c+3/d)=5/7 has 51 solutions? For example: (2,12,21,84) (2,5,141,29610) (8,8,8,14) or (2,9,20,945). Out of which 33 solutions are for a
@allenm935 3 жыл бұрын
Why can’t a b and c equal 4?
@Sergioqwe89 3 жыл бұрын
in the case where a = 3 there are 2 solutions b = 4 c = 6 and b = 3 c = 12
@rodrigo7026 3 жыл бұрын
What about a=b=c=4?
@cmilkau 3 жыл бұрын
Did you even watch? It's the very first solution mentioned, but there are more.
@kwankaewkanchanatam6876 3 жыл бұрын
(1/2)+(1/6)+(1/12)=3/4 So a=2, b=6, c=12 and a
@wit1729 3 жыл бұрын
3/4 - 1/3 = 5/12 and 28 is divisible by 7...
@tobiasgorgen7592 3 жыл бұрын
His solution still stands. First one was miss written but the inequality holds. Second one is 23/3 probably misspoken
@condorbz2 3 жыл бұрын
loves from Chile
@jorgegranada4964 3 жыл бұрын
I dont get it. If a+b+c =4 then its 2,1 ,1
@user-ht1vg5we2p 3 жыл бұрын
Me: 3/4- 1/3= 9/12 -4/12= 5/12 Michael: It EqUaLs SeVeN oVeR tWeLvE
@kajamix Жыл бұрын
28/7 not a natural number ?
@dhhgghggbhbbvfgg 3 жыл бұрын
Very cool
@mjackstewart 3 жыл бұрын
They’re all 4. Done!
@HungryEnigma79 3 жыл бұрын
So (a,b,c)=(4,4,4), (1/4)*3= 3/4 and c=b=a > 1
@wesley_tavares 3 жыл бұрын
It looks like some egyptian math problem
@matth8617 3 жыл бұрын
At 4:53, 3/4 - 1/3 is not 7/12, it’s 5/12. Thought I’d mention that
@matth8617 3 жыл бұрын
Also since 1/2 is more than 5/12, you could have moved forward with a = 3
@Dionisi0 3 жыл бұрын
28/7 isnt that 4?
@ffggddss 3 жыл бұрын
Yes. Only trouble is, 28/7 isn't what you get for that; it's 28/3, which isn't a natural number. Fred
@emmanueljansen1316 3 жыл бұрын
I am sorry but, why the solition isnt a = b = c = 4? There is no reason why you couldnt do that
@OOKIEDOKIE 3 жыл бұрын
I was thinking the same thing. It says "or equal to" so why not.
@teeweezeven 3 жыл бұрын
It is. That's what case 1 was for
@JB-ym4up 3 жыл бұрын
Question calls for all solutions.
@chandy3859 3 жыл бұрын
4:00 and the question did say "find all"
@emmanueljansen1316 3 жыл бұрын
Ah, thanks to all of you xD
@NihmbleTech 3 жыл бұрын
I see a video correction in the future.
@emanuelramirez9836 3 жыл бұрын
This problem would be a great coding exercise
@NoisqueVoaProduction 3 жыл бұрын
I know this one! You can make a=2,b=5 and c=20 I just made a test that I had to prove that a unit fraction can be divided as: 1/n=1/(n+1)+1/n(n+1) (Proof: 1/n=1/(n+1)+1/a 1/n=[a+(n+1)]/(n+1)a an+a=an+n²+n) So, a=2 makes 1/b+1/c=1/4 So n=4, b=n+1=5, c=n²+n=20 It is also pretty simple to see that 0.75=0.5+0.2+0.05
@kenichimori8533 3 жыл бұрын
=0a = 0b = 0c = a/b = b/a = c/a = a/c = b/c = c/b = 1 combination metricies.
@matthewsheeran 3 жыл бұрын
a=b=c=4; a=2 b=c=8; .. In my case would be quicker and easier to program a computer in python code to iteratively or even randomly check this in less than a dozen lines of code and this is in fact an acceptable algorithmic proof too!
@chandy3859 3 жыл бұрын
The question in the thumbnail and the question being solved is a bit different. Edit: read the question in the video before commenting
@TechToppers 3 жыл бұрын
I love the way he works out! (First time witnessing this method) Here is a quicker alternative: Consider 1/a + 1/b = 1/x Cross multiply ax+bx-ab=0 ab-ax-bx=0 a(b-x)-bx+x²=x² (a-x)(b-x)=x² Remember this, if preparing for Contest Math. Consider a simple version of problem: 1/a + 1/b = 3/4 1/3a + 1/3b =1/4 Well, use what we derived... (3a-4)(3b-4)= 16 Now you know what to do... 16= 1×16, 2×8, 4×4 You could reverse things, but I'm lazy... Start equating. First case not possible. 2nd case 3a-4=2 => a=2, b=4 3rd case is not possible... Now come to original problem: 1/a + 1/b + 1/c = 3/4 We know, 1/2+1/4= 3/4 How to proceed? Hmm... On thinking we find if we could split 1/2 as two unit fractions(you know what are they, right?) Then we get a solution? Right? Similarly split 1/4 and you're done. Remember that condition 1≤a≤b≤c. Rest you can as you already know, I'm lazy. This was very easy problem. A similar variant came in India for 5th graders in Olympiad. It's called NMTC. So I knew this by 💓!
@caesar_cipher 3 жыл бұрын
Your solution from 1/a+1/b=p/q with factorization (its a standard trick called SFFT) does not work for 1/a+1/b+1/c=p/q So not useful in this problem. Michael had done it in the shortest possible way - establish bounds on a, then b, then get c.
@TechToppers 3 жыл бұрын
@@caesar_cipher It should work out! Let me try...
@TechToppers 3 жыл бұрын
@@caesar_cipher My method, misses one solution...
@caesar_cipher 3 жыл бұрын
@@TechToppers Its not about missing 1 solution - the SFFT factorization does not work at all for 1/a+1/b+1/c=p/q
@TechToppers 3 жыл бұрын
@@caesar_cipher Please read the approach bro! I know you can not apply it *directly* . There is another neat trick to get solutions. As soon as we get 1/2 and 1/4, the strategy is to break 1/2 and 1/4 into unit fractions...
@rambot670 3 жыл бұрын
Looks like all the minor mistakes have been highlighted ! Great work comments section ! And good video Michael.
@sea34101 3 жыл бұрын
at 6:00 3/4-1/3 = 5/12 not 7/12 Hence there is the solution 1/3 + 1/4 + 1/6 = 3/4
@Straight_Talk 3 жыл бұрын
3/4 - 1/3 = 5/12 (at 5.00).
@maxspector7394 3 жыл бұрын
If they are all less than or equal to each other, couldn't they all be 4?
@ignaciorodriguez639 Ай бұрын
Great video By the way, I think this should be the work of a computer and not a human from fractions import Fraction from math import floor def q1(a, b, x): if (x.numerator == 1): c = x.denominator print('1 / %d + 1 / %d + 1 / %d = 3 / 4' % (a, b, c)) def q2(a, x): for b in range(max(a, floor(1 / x + 1)), floor(2 / x + 1)): q1(a, b, x - Fraction(1, b)) def q3(x): for a in range(floor(1 / x + 1), floor(3 / x + 1)): q2(a, x - Fraction(1, a)) q3(Fraction(3, 4))