pqr = 59 ( p + q + r)

This is a good question for practicing primality and algebra. Shout out to Nate W. for sending me both the problem and his solution. The solution you see in this video was provided along with the problem. I could not produce a more efficient solution.

[JoachimFavre]

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.

[maxgoldman]

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.

[myaccount]

(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.

[a_man]

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)

[Grecks]

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 <= r. Then 60 = (q-1)(r-1) >= (q-1)^2, therefore q-1 <= 7. So q must be in {2, 3, 5, 7} since it is a prime number. That greatly reduces the options. Let's consider each case:
4) q=2 implies r=61 (which is prime), that gives us (p, q, r) = (59, 2, 61) as a solution.
5) q=3 implies r=31 which yields the solution (59, 3, 31).
6) q=5 implies r=16 (which is composite), so we have no solution in this case.
7) q=7 implies r=11 which yields yet another solution (59, 7, 11).
8) In summary: There are 3 distinct (p, q, r) solution triples if we disregard the order of p, q, r. Any permutation of them is also a solution, so in total (respecting the order) we have 3 * 3! = 18 solutions.

[skwbusaidi]

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<r
And then get all premutations for p, q, r

[ProactiveYellow]

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.

[andreabaldacci]

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

[nirmalmishra]

5:50 add +1 and -1 on top and split the fraction then set r-1= all possible +ve factors of 60.

[emanuellandeholm]

Nice problem professor!

I got the WLOG, but my inference p <= q <= r, p or q or r = 59 was unproductive. You may as well set q = 59 with the same WLOG, because of symmetry. That's the thing I missed! :)

[swenpai]

You explain so well and clearly, thank you so much for all the videos!

[jasonryan]

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!❤

[pietergeerkens]

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.

[PrithwirajSen-njqq]

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)

[Muscisian-ny]

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.

[diogochadudmilagres]

[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 ✅

[Grecks]

"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. 😀

[AbhayPratapSinghRajawat-chuz]

It can also be done by Simon's favorite factorising trick and can be completed in just 2-3 min

[shacharh]

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.

[Dictr]

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!