Fish with binary speed

It is easy to see, that while the first (slowest) fish swims one round (back and forth), the second fish swims two rounds, the third fish 4 rounds ... and the nth fish $2^{n - 1}$ rounds.

Hence, all fish meets at the starting point whenever the first fish completes a round.

Therefore, our answer should be:

$\boxed { \text { Yes } }$

There would be a least common multiple of n quality.

$\text{fish}_{n}$ will have a speed of $2^{n} \cdot v_{1}$ , therefore we know that all fish has a common factor of speed, which (I think) implies that they will eventually meet up at the same end.

If you think of the distance of the pool as a unit then v1 travels 0.5 in the time v2 travels 1 and v3 travels 2. A fish will be at the right hand side of the pool on an odd number of units traveled and on the left side when an even amount. So if all fish can be at either an even or odd amount at the same time then it is possible.

Since each fish is twice the last, any time v1 is even the others are even. Ex: V1= 2 so V2=4 and V3=8

Their meeting time at starting point will be lowest common multiple of their individual time of covering a round.

One can deduce the answer without doing any maths. Clearly the answer cannot be "it depends if n is of our even" because if it works for n+1 fish that same value will work for n fish. Similarly it cannot depend on v because v is arbitrary and can only every be expressed in terms of tanks per time unit, in which case we just adjust our time units. Finally we are left with yes or no. Since it clearly works for n=2, the answer must be yes.

The slowest fish had gone half way when the other two had gone 1, and 2 laps. By the time the slowest fis has completed 2 laps, the other two wil have completed 4, and 8 laps respectively

(v1=v)=>(v2=2v)=>(v3=>4v).....

let at time t they are at same place again

for (fish one t=(2^a)x/v),(fish two t=(2^b)x/2v),...... (x=length between two ends)(a,b,.. are independent of each other)

for same time the so a=1,b=2,c=4,....

so yes its possible

It is easy to see that when 1st fish swims 1 round then 2nd fish swims 2 round, 3rd swims 4 rounds, 4th swims 8 rounds and so on. Hence at a particular all fishes will be at same end.

It's like rolling through the various different combinations on a combination lock but doubling the amount of twists you perform on each digit.

4 Digit Lock 0000, 1248, 2486, 3624, 4862, 5000, 6248, 7486, 8624, 9862, 0000 10 Moves back to zero

7 Digit Lock 0000000, 1248624, 2486248, 3624862, 4862486, 5000000, 6248624, 7486248, 8624862, 9862486, 0000000 10 Moves back to zero

They will be at the same end when the n-th fish has done 2^n one way trip, the n-1 th 2^(n-1) ... and the first (the slowest) 2 one way trip

Fish $1$ swims with speed $v_1$

Fish $2$ swims with speed $v_2 = 2 v_1$ ,

Fish $3$ swims with speed $v_3 = 2 v_2 = 4 v_1$ ,

Fish $4$ swims with speed $v_4 = 2 v_3 = 8 v_1$ ,

Fish $n$ swims with speed $v_n = 2 v_(n-1) = 2n v_1$ ,

$\therefore$ yes as $2v_1, 4 v_1, 8 v_1, 16 v_1..... 2n v_1 || v_1$