One can also use a different, but equivalent, reasoning for the end. pr = p + 59 + r pr - p - r + 1 = 60 (p-1)(r-1) = 60. Without loss of generality suppose r < p. We now look at the divisors of 60: If r-1 = 1 then p-1 = 60, which works. If r-1 = 2 then p-1 = 30, which works. If r-1 = 3 then p-1 = 20, which doesn't work. If r-1 = 4 then p-1 = 15, which doesn't work. If r-1 = 5 then p-1 = 12, which doesn't work. If r-1 = 6 then p-1 = 10, which works.
Very nice, it limits the possible candidates! I didn't get why Prime Newton knew that there wouldn't turn up any new primes that fit after trying 13 giving 6. EDIT: Of course, because p keeps falling, I get it now. But your method is much clearer.
I see. That is a nice identity you got. I just assume p = 59 then expressed q in terms of r as q=1+60/(r-1) so r-1 is a divisor of 60 but I missed on the symmetry of p and q and checked all possibilities
Since 59 is prime, either p, q, or r is 59. Let p1, p2 and 59 be the elements in the set {p, q, r}. Then p1*p2 = 59 + p1 + p2, which can be rewritten as (p1-1)(p2-1)=60. Factorize 60 and map two divisors in each possible factorization to p1-1 and p2-1 respectively. We reject the pairs (p1, p2) containing non-prime elements. This gives rise to (2, 61), (3, 31), (7, 11), (11, 7), (31, 3), (61, 2). Hence, all possible prime triples are (2, 59, 61), (3, 31, 59), (7, 11, 59), and the permutation of each one.
Given that 59 is a prime number, and because of how primes are defined, we know that two products are equal iff their prime factorizations are equal. Since 59 is prime, it must be p,q, or r. WLOG we can assign it to be p. By the symmetric nature, we can solve for q in terms of r, giving us that q=1+60/(r-1). We know q is a prime, and the set of all primes P is a strict subset of the natural numbers N, so it must be that 60/(r-1) is in N. This means that the possible values for r-1 are a subset of the factors of 60, specifically those which are one less than a prime. Let S be the set of the factors of 60, and S+1 be the {s+1| s in S} then the set of possible values for r is intersect(P,(S+1)). From here we can then define the set of solutions for q to be intersect(p,{k-1 | k in intersect(P,(S+1))}). After proper pairing of values, we get that the set {p,q,r} satisfying our conditions corresponds to the (unordered!) sets {2,59,61}, {3,31,59}, and {7,11,59}, and because of the symmetry of our conditions, the set of ordered triplets (p,q,r) satisfying our problem is the set of all permutations on those three sets, having a total of 3!×3=18 triplets.
What is said in the video can also be seen through the graph of the function p = (59 + r)/(r - 1) which is an equilateral hyperbola with an axis of symmetry p = r and with vertex (I neglect the part that has negative p or r) the point that has coordinates p = 8,... r = 8,... therefore it makes no sense to give r values higher than 7 and you will essentially find the three sets of values indicated in the video.
My solution was like that: after pr=p+r+59 , pr-(p+r)=59 I analyzed divisibility by 3 of p and r. 1)p=r=3 2)p=3 (same as r=3 (actually case 1 is also trivial)) 3)p=3k+1 r=3n+1 4)p=3k+1 r=3n+2 5)p=3k+2 r=3n+2 there is no case something like p=3 r=3k+1 or p=3 r=3k+2 because the equation already gives you the value of r, if it is prime then it's a solution, if it's not then not a solution. I won't write the analysis of cases because it is a little bit of too long (especially case 3, 4 and 5)
You can modify your last equation to limit your choices of r p= (r+59)/(r-1)=(r-1+60)/(r-1) p= 1 + 60/(r-1) We only think of prime number r that one more than factor of 60 It is eaiser if solve the diaphantine equation (p-1)(r-1)=60 and wlog assume p
My approach: after establishing that the porblem is symmetric and after picking p=59, semplification brings to qr-r=59+q. Then r(q-1)=59+q. If q is even (hence the only case is q=2) we get immediately r=61, which works. For q odd, we can write q=2a+1, and substituting we get 2ar=60+2a, or ar=30+a, which is a(r-1)=30. Trying out the factors of 30 for a and r-1 brings quickly tot he solution
My way to a solution: 1) I started in the same way as @PrimeNewtons: 59 is a prime number that divides the product pqr, so it must divide (at least) one of p, q, or r. Since those variables stand for primes themselves, (at least) one of them must be equal to 59. W.l.o.g. let that prime be p. (Note: The problem is highly symmetric; if one specific (p,q,r) triple is a solution, any other permutation is, too.) 2) We divide by p and get: qr = 59+q+r. After adding 1 on both sides and rearranging we get: (q-1)(r-1) = qr-q-r+1 = 59+1 = 60. 3) W.l.o.g. we can assume q = (q-1)^2, therefore q-1
Time slot 5-00 Here I like to solve as under pr = p +59+r >pr -p -r +1=60 >(p-1)(r-1)=60 If p-1= 1 then r-1=60 (p=2,r=61) p-1=2 then r-1=30(p=3 r=31) If p-1=3 p=4 not prime If p-1=4 then r-1=15 (p=5 r=16 16 is not prime) If p-1=5 then p=6 is not prime If p-1=6 then r-1=10 (p=7 r=11) If p-1=7 p=8 not prime If p-1=9 p=10 not prime If p-1=10 p=11 r-1=6 r=7-- now p and r are interchanging their prime values Hence the triples are (p, q, r) =(2,59,61) (3,59,31) (7,59,11)
To complete the answer: all 6 solutions are (p,q,r): (2,59,61); (61,59,2); (3,59,31); (31,59,3); (7,59,11) and (11,59,7). At r=7 there is symmetry, so r can be bigger than 7, yet only the former found values of p.
"You pick the letter you don't want to keep writing. So which one is the least attractive?" LOL. Man, I love your efficiency. That could have been me. And I agree, "q" is the least attractive one. Nevertheless, I chose "p" because it was the first variable. 😀
Simon's factorizing trick It can be used when Diophantine equation of integers such as axy+bx+cy = d assuming a,b,c,d,x,y are integers, axy+bx+cy = d => x(ay+b)+c/a(ay+b) = d+bc/a (ay+b)(x+c/a) = d+bc/a => (ay+b)(ax+c) = ad+bc pr = p + 59 + r => pr - p - r = 59 p(r -1) - (r -1) = 60 => (p -1)(r -1) = 60.
I'd have to like this video. The property of associative and commutative functions was so useful in this specific question. I loved it, thank you Mr. Newton's!❤
At 10:20 you mistakenly suggested that if you keep going beyond r=7, you will get numbers "that are not prime ... or not prime". You meant to say "that are not prime, or have already been found". For example, r=31 gives p=3, and r=61 gives p=2. However, the expression (r+59)/(r-1) is decreasing and asymptotic to 1 as r increases, and is less than 2 for r > 61. That guarantees that no values greater than 61 can be part of any solution.
[4:32] pr = p + 59 + r, or PRODUCT = SUM + 59, or PRODUCT - SUM = 59 >> p.r - (p+r) = 59. This says something about Girard relations. But i don't know use this in this problem, maybe is useful to think in prime numbers such that product are greater than 59, like 7 and 11... 7*11 - (7+11) = 77 - 18 = 59 ✅
Okay that last part wasn't that rigorous so here's the technical approach: r-1 must divide 59+r because p is an integer so r-1|59+r but r-1|r-1 which means r-1|59+r-(r-1) ==> r-1|60 therefore r-1 is a divisor of 60.
You can prove analytically that yours are the only 3 solutions. as you raise r, (r+59)/r-1) tends to 1 at the limit. So take epsilon < 1, then for some N if r>N: (r+59)/r(+1) -1 < epsilon < 1. If you continue this way, one N that works is 61. So in particular for every prime above 61, this means (r+59)/(r-1) < 2. Which then means the triplets can't contain primes bigger than 61.
Nice! And thanks to Nate W. as well. I brute-forced (using Viete after identifying 59) the solution from the quadratic x^2 - s * x + 59 + s = 0 which, upon setting the discriminant equal to n^2, gives in turn (s - 2)^2 - n^2 = 4 + 4*59 = 4 + 236 = 240. Then gnomon analysis generates all the correct solutions as well as a few non-satisfying ones. This method, though more tedious, DOES however answer the question "Why 59?": Because, in addition to being prime, 59 * 4 + 4 = 240, which has LOTS of even factors.
This ins't rigorous enough From pr = p + 59 + r you can add one to both sides and do some algebra to factorize it as 60 = (p-1) (q-1) from then you can check all the factor pairs of 60 and conclude that those are the only solutions
@@KingshipKRP His method is essentially to assume that the problem makers would not make an absurdly high prime part of one of the solutions. While fine for a problem like this, if the question was "find all primes p, q, r s.t... AND prove that these are the only prime triplets", we would need to do more work.
Interesting problem and solution. Would be interesting to see what other primes can be used in place of 59 and which , if any, primes lead to more than 3 solutions ?
I really like how these types of equations are not solved in my conventional way of solving them. I don't typically use "logic" I just solve by numbers 😁. Nice to see that equations can be solved by more than just numbers. 😃 Wonderful, keep it up!
The only problem is that none of the ‘answers’ are prime triples which must have the form (p, p+2, p+6) or (p, p+4, p+6). So while your working out is neat and effective, it is actually answering a different question. Here you are simply searching for disconnected primes that fit the bill. So the issue is nomenclature. The problem as stated is not the problem answered.
Instead of 59 try 97. Any solutions? How about 29? What if you try the smaller of a pair of consecutive primes? eg. 59,61 or 29,31 unlike 97,99 p=1 + (59 + 1)/(r - 1) r = 61 guarantees 1 solution and similarly for 29,31. Not always solutions for arbitrary q , q + 2 with q prime. Bit of mileage still in the problem. For what primes k does ab = a + b + k have solutions with a , b prime is pretty much a rewrite of the original problem where essentially q was fixed arbitrary 59? prime.
p = (59+r)/(r-1) = (60+r-1)/(r-1) = 1 + 60/(r-1) As r increases p decreases. We try all primes less than 60. Alternatively, can try factors of 60 too, but it'll be lengthier ig.
@@nanamacapagal8342 60 = 2×2×3×5. Now the combinations of these 4 may be more than primes under 60. Plus you'd have to think. Edit: Ig it's equivalent.
I was recently at a math camp where we were given problems to solve each day and the younger group (which I wasn't in) got this exact problem once. I wonder where this is from (I haven't watched the video yet).
A prime is any (non-unit) number p such that if p divides a product mn, p must divide one of the factors m or n. We know 59 is prime, and it divides pqr, so by the definition of primes above, 59 divides p, or it divides qr. 1. If it divides p, and p is prime, then it must be equal to p, since p only has factors of 1 and itself 2. If 59 divides qr, then by the definition above we again know that 59 divides q or it divides r. In both cases, these are prime numbers, so just like in case 1, one of them must be equal to 59.
@@ProactiveYellow Exactly. The first sentence is the definining property of *prime element* in any commutative ring (in addition to being non-zero, non-unit elements). In such general rings "prime" is *not* the same notion as "irreducible" (i.e. cannot be non-trivially factored) but in integral domains at least, the former implies the latter.
Old title/mistake sol pqr=58(p+q+r) pqr=29×2(p+q+r) as 29 and 2 two are prime and the three numbers are prime either p q or r is 29 or 2 Wlog p=29 and q=2 r=31+r 0=31 No sols
This is an "intriguing" attempt to drag traffic from this site to your tutorial site. You got into the game way too late. Math help web sites exist in excessive abundance and the competition is fierce.
I didn't expect from you to give such a bad solution to such a simple problem. (r+59)/(r-1)=1+60/(r-1). This is a classic divisibility problem. So (r-1) must divide 60 and from there all the solutions will be found. How to conclude that, if 13 is not good, no other prime number bigger than 13 will be good?! It's totally wrong.
I think the problem here is that what you typed makes sense to you. Do you think it makes sense to anyone else reading it? The fact that you can find the answer does not mean others want to do it your way. Just a suggestion for next time, say your solution and let others say they like it.
@@PrimeNewtons I have already said my solution and it is already presented by many others. Your solution is wrong. You cannot make a generalized, unproven statement.
What I think he meant to say right at the end when he said "they are either not prime or not prime", was that because the value of r had been falling as the value of p increased, and because with the p=13 example (which was totally not chosen randomly as he claimed) the value of r had fallen to 6, it means that the only prime values that r could be after that are 5,3 and 2, but we already found the solutions for p=2, p=3 and the invalid solution for p=5, so we'd only be finding the same solutions again but with p and r switched, therefore we have found all the solutions by that stage. I can only assume it was either a very hot day, or else it was just before lunch and his blood sugar dipped by the end of the recording session, as he sort of lost track of his own conclusion at the end, but we're all human and such things happen to us all.
Hello... This trying all primes for r is my problem!!! You are using brute force in deciphering numbers.. Nothing new.. Wait for me i just come to maths world. I will find solutions trust me😂
One can also use a different, but equivalent, reasoning for the end.
pr = p + 59 + r pr - p - r + 1 = 60 (p-1)(r-1) = 60.
Without loss of generality suppose r < p. We now look at the divisors of 60:
If r-1 = 1 then p-1 = 60, which works.
If r-1 = 2 then p-1 = 30, which works.
If r-1 = 3 then p-1 = 20, which doesn't work.
If r-1 = 4 then p-1 = 15, which doesn't work.
If r-1 = 5 then p-1 = 12, which doesn't work.
If r-1 = 6 then p-1 = 10, which works.
Very nice, it limits the possible candidates!
I didn't get why Prime Newton knew that there wouldn't turn up any new primes that fit after trying 13 giving 6. EDIT: Of course, because p keeps falling, I get it now. But your method is much clearer.
I think this solution is highly versatile so when you come across another similar question,you can try the same way.
Good solution! But you don't need to go through all divisors r-1 ≤ 6 because r must be prime. So r-1=3 and r-1=5 are not to be checked.
I see. That is a nice identity you got. I just assume p = 59 then expressed q in terms of r as q=1+60/(r-1) so r-1 is a divisor of 60 but I missed on the symmetry of p and q and checked all possibilities
I was surprised this was not used in the video since we have used similar factorizations in the past for similar problems.
Since 59 is prime, either p, q, or r is 59. Let p1, p2 and 59 be the elements in the set {p, q, r}. Then p1*p2 = 59 + p1 + p2, which can be rewritten as (p1-1)(p2-1)=60. Factorize 60 and map two divisors in each possible factorization to p1-1 and p2-1 respectively. We reject the pairs (p1, p2) containing non-prime elements. This gives rise to (2, 61), (3, 31), (7, 11), (11, 7), (31, 3), (61, 2). Hence, all possible prime triples are (2, 59, 61), (3, 31, 59), (7, 11, 59), and the permutation of each one.
(p-1)(r-1) = 60 = 2 • 2 • 3 • 5.
Now group factors in all possible ways such that their product is one unit less than a prime.
Nice, but you forget the factor 1
Given that 59 is a prime number, and because of how primes are defined, we know that two products are equal iff their prime factorizations are equal. Since 59 is prime, it must be p,q, or r. WLOG we can assign it to be p. By the symmetric nature, we can solve for q in terms of r, giving us that q=1+60/(r-1). We know q is a prime, and the set of all primes P is a strict subset of the natural numbers N, so it must be that 60/(r-1) is in N. This means that the possible values for r-1 are a subset of the factors of 60, specifically those which are one less than a prime. Let S be the set of the factors of 60, and S+1 be the {s+1| s in S} then the set of possible values for r is intersect(P,(S+1)). From here we can then define the set of solutions for q to be intersect(p,{k-1 | k in intersect(P,(S+1))}). After proper pairing of values, we get that the set {p,q,r} satisfying our conditions corresponds to the (unordered!) sets {2,59,61}, {3,31,59}, and {7,11,59}, and because of the symmetry of our conditions, the set of ordered triplets (p,q,r) satisfying our problem is the set of all permutations on those three sets, having a total of 3!×3=18 triplets.
What is said in the video can also be seen through the graph of the function
p = (59 + r)/(r - 1)
which is an equilateral hyperbola with an axis of symmetry
p = r
and with vertex (I neglect the part that has negative p or r)
the point that has coordinates
p = 8,... r = 8,...
therefore it makes no sense to give r values higher than 7 and you will essentially find the three sets of values indicated in the video.
My solution was like that: after pr=p+r+59 , pr-(p+r)=59
I analyzed divisibility by 3 of p and r.
1)p=r=3
2)p=3 (same as r=3 (actually case 1 is also trivial))
3)p=3k+1 r=3n+1
4)p=3k+1 r=3n+2
5)p=3k+2 r=3n+2
there is no case something like p=3 r=3k+1 or p=3 r=3k+2 because the equation already gives you the value of r, if it is prime then it's a solution, if it's not then not a solution. I won't write the analysis of cases because it is a little bit of too long (especially case 3, 4 and 5)
Nice problem professor!
I got the WLOG, but my inference p
You can modify your last equation to limit your choices of r
p= (r+59)/(r-1)=(r-1+60)/(r-1)
p= 1 + 60/(r-1)
We only think of prime number r that one more than factor of 60
It is eaiser if solve the diaphantine equation (p-1)(r-1)=60 and wlog assume p
p=1 + 60/(r-1)
My approach: after establishing that the porblem is symmetric and after picking p=59, semplification brings to qr-r=59+q. Then r(q-1)=59+q. If q is even (hence the only case is q=2) we get immediately r=61, which works. For q odd, we can write q=2a+1, and substituting we get 2ar=60+2a, or ar=30+a, which is a(r-1)=30. Trying out the factors of 30 for a and r-1 brings quickly tot he solution
Excellent approach
My way to a solution:
1) I started in the same way as @PrimeNewtons: 59 is a prime number that divides the product pqr, so it must divide (at least) one of p, q, or r. Since those variables stand for primes themselves, (at least) one of them must be equal to 59. W.l.o.g. let that prime be p. (Note: The problem is highly symmetric; if one specific (p,q,r) triple is a solution, any other permutation is, too.)
2) We divide by p and get: qr = 59+q+r. After adding 1 on both sides and rearranging we get: (q-1)(r-1) = qr-q-r+1 = 59+1 = 60.
3) W.l.o.g. we can assume q = (q-1)^2, therefore q-1
Time slot 5-00
Here I like to solve as under
pr = p +59+r
>pr -p -r +1=60
>(p-1)(r-1)=60
If p-1= 1 then r-1=60 (p=2,r=61)
p-1=2 then r-1=30(p=3 r=31)
If p-1=3 p=4 not prime
If p-1=4 then r-1=15
(p=5 r=16 16 is not prime)
If p-1=5 then p=6 is not prime
If p-1=6 then r-1=10 (p=7 r=11)
If p-1=7 p=8 not prime
If p-1=9 p=10 not prime
If p-1=10 p=11
r-1=6 r=7-- now p and r are interchanging their prime values
Hence the triples are
(p, q, r) =(2,59,61)
(3,59,31)
(7,59,11)
5:50 add +1 and -1 on top and split the fraction then set r-1= all possible +ve factors of 60.
To complete the answer: all 6 solutions are (p,q,r): (2,59,61); (61,59,2); (3,59,31); (31,59,3); (7,59,11) and (11,59,7). At r=7 there is symmetry, so r can be bigger than 7, yet only the former found values of p.
"You pick the letter you don't want to keep writing. So which one is the least attractive?" LOL.
Man, I love your efficiency. That could have been me. And I agree, "q" is the least attractive one. Nevertheless, I chose "p" because it was the first variable. 😀
You explain so well and clearly, thank you so much for all the videos!
It can also be done by Simon's favorite factorising trick and can be completed in just 2-3 min
I want to know Simon's trick
@@PrimeNewtons you can see it in the solution of
@JoachimFavre
Yes, me too. Please show us.
Simon's factorizing trick
It can be used when Diophantine equation of integers such as axy+bx+cy = d
assuming a,b,c,d,x,y are integers,
axy+bx+cy = d => x(ay+b)+c/a(ay+b) = d+bc/a
(ay+b)(x+c/a) = d+bc/a => (ay+b)(ax+c) = ad+bc
pr = p + 59 + r => pr - p - r = 59
p(r -1) - (r -1) = 60 => (p -1)(r -1) = 60.
@@9허공 Great, thanks. Is that Diophantine equation (axy+bx+cy = d) the most general form where Simon's factorization trick can be applied?
I'd have to like this video. The property of associative and commutative functions was so useful in this specific question. I loved it, thank you Mr. Newton's!❤
At 10:20 you mistakenly suggested that if you keep going beyond r=7, you will get numbers "that are not prime ... or not prime". You meant to say "that are not prime, or have already been found". For example, r=31 gives p=3, and r=61 gives p=2.
However, the expression (r+59)/(r-1) is decreasing and asymptotic to 1 as r increases, and is less than 2 for r > 61. That guarantees that no values greater than 61 can be part of any solution.
[4:32] pr = p + 59 + r, or PRODUCT = SUM + 59, or PRODUCT - SUM = 59 >> p.r - (p+r) = 59. This says something about Girard relations. But i don't know use this in this problem, maybe is useful to think in prime numbers such that product are greater than 59, like 7 and 11... 7*11 - (7+11) = 77 - 18 = 59 ✅
Okay that last part wasn't that rigorous so here's the technical approach:
r-1 must divide 59+r because p is an integer so r-1|59+r but r-1|r-1 which means r-1|59+r-(r-1) ==> r-1|60 therefore r-1 is a divisor of 60.
P = (59 + r)/(r-1) >= 2 ;
59+r >= 2(r-1) = 2r -2 ;
59 +2 >= 2r - r = r ;
Therefore, r
You can prove analytically that yours are the only 3 solutions.
as you raise r, (r+59)/r-1) tends to 1 at the limit.
So take epsilon < 1, then for some N if r>N: (r+59)/r(+1) -1 < epsilon < 1.
If you continue this way, one N that works is 61.
So in particular for every prime above 61, this means (r+59)/(r-1) < 2.
Which then means the triplets can't contain primes bigger than 61.
PLEASE solve more JEE ADVANCED Questions.
I am an jee aspirant, i find your videos very helpful i have learnt many skills from your videos.
typo in video title
Yes. There is a mistake in the title.
Up
not anymore
Nice! And thanks to Nate W. as well.
I brute-forced (using Viete after identifying 59) the solution from the quadratic
x^2 - s * x + 59 + s = 0
which, upon setting the discriminant equal to n^2, gives in turn
(s - 2)^2 - n^2 = 4 + 4*59 = 4 + 236 = 240.
Then gnomon analysis generates all the correct solutions as well as a few non-satisfying ones.
This method, though more tedious, DOES however answer the question "Why 59?":
Because, in addition to being prime, 59 * 4 + 4 = 240, which has LOTS of even factors.
This ins't rigorous enough
From pr = p + 59 + r
you can add one to both sides and do some algebra to factorize it as 60 = (p-1) (q-1)
from then you can check all the factor pairs of 60 and conclude that those are the only solutions
Thanks!
Quite elegant! 👍
You got the same result, how is the method in the video not rigorous enough?
@@KingshipKRP His method is essentially to assume that the problem makers would not make an absurdly high prime part of one of the solutions. While fine for a problem like this, if the question was "find all primes p, q, r s.t... AND prove that these are the only prime triplets", we would need to do more work.
agree, this is easy. i used the same method as you!
Interesting problem and solution. Would be interesting to see what other primes can be used in place of 59 and which , if any, primes lead to more than 3 solutions ?
I really like how these types of equations are not solved in my conventional way of solving them. I don't typically use "logic" I just solve by numbers 😁. Nice to see that equations can be solved by more than just numbers. 😃 Wonderful, keep it up!
(p-1)(r-1)=60 by adding 1 to both sides. Then analyzing the prime factors of 60 is trivial.
The only problem is that none of the ‘answers’ are prime triples which must have the form (p, p+2, p+6) or (p, p+4, p+6). So while your working out is neat and effective, it is actually answering a different question. Here you are simply searching for disconnected primes that fit the bill. So the issue is nomenclature. The problem as stated is not the problem answered.
Instead of 59 try 97. Any solutions? How about 29? What if you try the smaller of a pair of consecutive primes? eg. 59,61 or 29,31 unlike 97,99
p=1 + (59 + 1)/(r - 1) r = 61 guarantees 1 solution and similarly for 29,31. Not always solutions for arbitrary q , q + 2 with q prime. Bit of mileage still in the problem.
For what primes k does ab =
a + b + k have solutions with a , b prime is pretty much a rewrite of the original problem where essentially q was fixed arbitrary 59? prime.
Couldn't you express the general solution in terms of the set {p,q,r} and a set of sets?
p = (59+r)/(r-1)
= (60+r-1)/(r-1)
= 1 + 60/(r-1)
As r increases p decreases. We try all primes less than 60. Alternatively, can try factors of 60 too, but it'll be lengthier ig.
Aren't there less factors of 60 than primes under 60?
@@nanamacapagal8342 60 = 2×2×3×5. Now the combinations of these 4 may be more than primes under 60. Plus you'd have to think.
Edit:
Ig it's equivalent.
Find all prime triples such that pqr=59(p+q+r) (61,59,2)(31,59,3) and (11,59,7) It’s in my head.
Are you a genius
@@Generalist18 Yes I am😎❤️💙🩵
Brotha that was smooth as hell😂
bro didnt use simons favorite factoring trick :cry:
59,61 are prime.Difference 2 is also prime.
So p=2,q=59,r=61
10:40 --- ❤️❤️❤️
amen brother : )
(59+3)/(3-1)=31
Sir this might be an unrelated question but what is ur personal favorite fields of maths ?
What about negative prime numbers
Are you not going to try them
I was recently at a math camp where we were given problems to solve each day and the younger group (which I wasn't in) got this exact problem once. I wonder where this is from (I haven't watched the video yet).
I like your handwriting.
Ι didn't understand why one of the variables MUST be 59 can someone explain in more detail ??
We have to find the prime number and in prime number, it's divisible by it's own no. or 1 so that's it
A prime is any (non-unit) number p such that if p divides a product mn, p must divide one of the factors m or n.
We know 59 is prime, and it divides pqr, so by the definition of primes above, 59 divides p, or it divides qr.
1. If it divides p, and p is prime, then it must be equal to p, since p only has factors of 1 and itself
2. If 59 divides qr, then by the definition above we again know that 59 divides q or it divides r. In both cases, these are prime numbers, so just like in case 1, one of them must be equal to 59.
@@ProactiveYellow Exactly. The first sentence is the definining property of *prime element* in any commutative ring (in addition to being non-zero, non-unit elements). In such general rings "prime" is *not* the same notion as "irreducible" (i.e. cannot be non-trivially factored) but in integral domains at least, the former implies the latter.
Lets choose a random prime number after 11....
Chooses next one in sequence.
1st to learn sir😁
Old title/mistake sol pqr=58(p+q+r)
pqr=29×2(p+q+r)
as 29 and 2 two are prime and the three numbers are prime either p q or r is 29 or 2 Wlog p=29 and q=2
r=31+r
0=31
No sols
29x2=58, not 59.
(59+7)/(7-1)=11
There is a fallacy in your “hit or miss” method. The strict procedures for finding all triplets has been suggested above.
Why haven't you uploaded a video about beta function 😢
You are amazing
The title of video should be changed. There is 58 in lieu of 59
Thanks. Done.
Number theory : one of the most intriguing subjects in math.
This is an "intriguing" attempt to drag traffic from this site to your tutorial site.
You got into the game way too late. Math help web sites exist in excessive abundance and the competition is fierce.
I didn't expect from you to give such a bad solution to such a simple problem.
(r+59)/(r-1)=1+60/(r-1). This is a classic divisibility problem.
So (r-1) must divide 60 and from there all the solutions will be found.
How to conclude that, if 13 is not good, no other prime number bigger than 13 will be good?!
It's totally wrong.
I think the problem here is that what you typed makes sense to you. Do you think it makes sense to anyone else reading it? The fact that you can find the answer does not mean others want to do it your way. Just a suggestion for next time, say your solution and let others say they like it.
@@PrimeNewtons I have already said my solution and it is already presented by many others.
Your solution is wrong.
You cannot make a generalized, unproven statement.
Case settled.
❤️
So if 13 does not work, that means we have found all of the solutions? Is this a proof that we have all solutions?
The last part of the solution wasn't rigorous at all.
What I think he meant to say right at the end when he said "they are either not prime or not prime", was that because the value of r had been falling as the value of p increased, and because with the p=13 example (which was totally not chosen randomly as he claimed) the value of r had fallen to 6, it means that the only prime values that r could be after that are 5,3 and 2, but we already found the solutions for p=2, p=3 and the invalid solution for p=5, so we'd only be finding the same solutions again but with p and r switched, therefore we have found all the solutions by that stage. I can only assume it was either a very hot day, or else it was just before lunch and his blood sugar dipped by the end of the recording session, as he sort of lost track of his own conclusion at the end, but we're all human and such things happen to us all.
17, 19 and 23
okok, eu aceito que preciso estudar muitíssimo mais
q = 59, but in general
q = 59.k.
Diofantina solução
Hello...
This trying all primes for r is my problem!!!
You are using brute force in deciphering numbers..
Nothing new..
Wait for me i just come to maths world.
I will find solutions trust me😂
Нашёл
Hi
P=(59+2)/(2-1)=61
(59+3)/(3-1)=31
(59+7)/(7-1)=11
(61,59,2),(31,59,3) and (11,59,7) It’s in my head.
Hi
(59+2)/(2-1)=61