# Can you solve the fish riddle? – Steve Wyborney

View full lesson: http://ed.ted.com/lessons/can-you-solve-the-fish-riddle-steve-wyborney

As the cargo director on the maiden voyage of the S.S. Buoyant, you’ve agreed to transport several tanks containing the last specimens of an endangered fish species to their new aquarium. Unfortunately, the boat is battered by a fierce storm, throwing your precious cargo overboard. Can you get the fish to safety and save the day? Steve Wyborney shows how.

Lesson by Steve Wyborney, animation by Artrake Studio.

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hardest riddle?!

If you had enough fuel to search all three sectors, why would it matter how many tanks there are?

This one isn't that hard like the other one that I watched earlier..

I just assumed that if I multiplied the 13 tanks by 2 (2 fishes per tank) so it would be 26 and it would leave us 24 sharks and saying that here is already 6 sharks it would leave us more than 7 sharks in the gamma.

So I assume that there are 3 fishes per tank

And I just computed it and I was right.

Basically I got it by Try and Error but I was lucky that I got it in 1st try.

(Someone might say its second but I already thought about the 2 fishes per tank that it wouldn't be enough to have 50 organisms)

So yeah.. I hope someone believed me :3

I FIGURED THIS ONE!! FIRST RIDDLE OF THE NIGHT, I AM NOT BRAIN DEAD

this is the first one in this riddle series that I can solve. Feel great

I believe that the answer provided, 13 tanks in total is not the only answer.

One cannot assume that the total number of sharks is 7. In fact, the shark number is from 7 ( 6 + 1 ) to 13 ( 6 + 7).

That said, the total number of fishes has a range of 43 to 37. There are 13 or fewer tanks in total. With the 6 tanks found, the range of tanks is 13 to 6. Hence, you result in 44<xy<36 and 14<y<5, where x and y are both whole numbers. (If you don't know why I used 44, 36, 14 and 5, I'm sorry I couldn't find inequalities with a line under it so I used this to substitute for what was needed).

Out of the numbers, you get

42 (7×6) (6×7)

40 (10×4) (8×5)

39 (13×3)

To prove that they are all correct, let's analyse each answer separately.

42 rare fishes in total, with 7 tanks in total and 6 fishes each. Sharks = 8 (2+4+2).

42 rare fishes in total, with 6 tanks in total and 7 fishes each. Sharks = 7 (2+4+1).

40 rare fishes in total, with 10 tanks in total and 4 fishes each. Sharks = 10 (2+4+4).

40 rare fishes in total, with 8 tanks in total and 5 fishes each. Sharks = 10 (2+4+4).

And lastly,

39 rare fishes in total, with 13 tanks in total and 3 fishes each. Sharks = 11 (2+4+5).

Cancerous.

There's usually many more fish than just that population meaning not all the fish would be killed allowing the ones that escape to breed causing it to make it a bit easier on you, also sharks are usually very picky enless they're bull sharks or tiger sharks which gives you other advantages, then you also have the case that they're made of a metal like mineral and if you were smart you probably brought your calculator, meaning it's much more simple than it seam, other advantages includ you probably fed the fish before this happened and that that means they should not be effected for about 12 hours, oh, and sharks will fight each other if they're not the same species.

The only place where I got messed up was, he said there was slots for tanks upto 13 tanks. Which means it could've been lower too

But then he just assumed all the freaking slots are taken and 7 tanks were in sector gamma. That's not fair, by my opinion

Why can't you just go through all three and pick up as many fish as possible until you run out of fuel?

Too many rules I can do it.

I used algeba instead of a table but got the same answer so its fine

i solved it!

Im correct!

And I thought stubbing my toe on furniture was bad luck😂

there is only 6 fish in alpha and gamma shown before altogether

Whooh! I actually got this one right!

The fact that I managed to solve this but couldn't figure out the virus one is just sad

13 tanks altogether

3 fish a piece

That's 39

Plus 2+4+5sharks

That's correct unless this is one of those multiple answer ones let's aee

Ok this captain has no idea what his ship is carrying? Boo

are these fish the same species as the fish in Einstein's riddle?

I always solve these riddles with Excel, I was surprised how easy it was this time with it, and even more when the answer was asking us to use a table!

I did it slightly differently than the solution. 7 to 13 on X and Y axis and =(50-$A2)/B$1 copied in all direction, keep only the rounded numbers. Delete 8 and 10 sharks rows and there you have it!

My solution :

( 4T + 2 ) + ( 2T + 4 ) + ( nT + m ) = 50 ( T->no of fish in each tank, n -> no of tanks in gamma, m ->no of sharks in gamma )

=> (6+n)T + (m+6) = 50;

now,

m can't be 2 and 4; {which leaves us with 1,3,5,6,7 }

m can't be 1 because, if it is one, we get (6+n)T = 43 but 43 is a prime number;

m can't be 3, … same reason, 41 is prime,

m can't be 7,…. same reason, 37 is prime

m can be 6 but, it leaves us with (6+n)T = 38 (w.k.t only 2*19 = 38) and => n=13 which is not possible because total is 13 .

So m is 5, i.e, (6+n)T=39 =3*13 => T=3 , n =7 😀

this was so easy. and that's saying something cause i never get your guys' riddles.

That was easy.

I came up with another solution, there could be 2 sharks in Gamma Sector and 1 tank. This means,

(2+4x)+(4+2x)+(2+x)= 50

=7x+8=50

=7x=42

=x=6

So there also could be 6 fish in each tank and be a total of 7 tanks and 8 sharks.

I actually got it right 0.0. Trial and error I love you XD

You lost me at the captain telling him there's a "strange property" in the water where no two sectors have the same amount of sharks, less than one shark, or more than seven. Are you even vaguely trying to be realistic?

ME: This guy is more unlucky than me!