# Challenge .....(Long One)?

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Peter, Paul and Anthony are honest fishermen. They went out to the ocean to catch fish on a small boat.They had a good catch. By the time they returned, it was nighfall. It is an old fisherman's saying "never split your fish after sunset". So, they decide to wait till daybreak.They make small tents with palm leaves for each of them and go to sleep in them.

It's next day morning.

Peter gets up at 5:30am and sees the sun rising but does not see the others out of their tents. Not wanting disturb their sleep, he goes to the catch and counts the number of fish they had caught the previous day.The count is not exactly divisible by 3. To divide equally and avoid confusion, he has to throw one fish back into the ocean. Like a gentleman he does that, he takes one-third and leaves.

Paul gets up at 6:00am and sees it is morning but does not see the others out of their tents (Yes, Peter left his empty tent standing). Assuming no one has got up from their sleep, he goes to the catch and counts the number of fish they had caught the previous day.The count is not exactly divisible by 3. To divide equally and avoid confusion, he has to throw one fish back into the ocean. Like a gentleman he does that, he takes one-third and leaves.

Anthony gets up at 6:30am and sees it is morning but does not see the others out of their tents (Yes, Peter and Paul have left their empty tents standing). Assuming no one has got up from their sleep, he too goes to the catch and counts the number of fish they had caught the previous day. The count is not exactly divisible by 3. To divide equally and avoid confusion, he has to throw one fish back into the ocean. Like a gentleman he does that, he takes one-third and leaves.

Now, the questions.

1. What is the smallest possible number of fish they could have caught so that, this mathematical calculation by all three fishermen is possible?

2. What is the remainder after Peter, Paul and Anthony have taken their "One-thirds"?
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1 I reckon the total number of fish is 79 ( First man thows one back to leave 78 divide by 3 = 26. Take 26 from 78 to leave 52. Second man throws one back to leave 51 divide by 3 = 17. Take 17 from 51 leaves 34. Third man throws one gack to leave 33 divide by 3= 11)

2 So the remaining fish is 22 ( 33 - 11)
Eargasms, Good!

But you can better that answer.
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Would the lowest remainder of fish be two? (I don't claim to be good at math)

The total catch be forty?
No Benita.

Starting with 40....throw away 1, becomes 39

1/3rd taken, leaves 26 behind,

throw away 1, becomes 25 (and still not divisible by 3).
giggle giggle - I know! (Do you really think I'm that bad at math??)

You would make a good professor - but ........what would be your subject??

By the way - this mathematical riddle/puzzle has a proper set of equations. Why don't we just let others answer.
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Hi FluidnMotion

Here's my latest effort......................

1/3rd taken leaves 34 throw one away 33

1/3rd taken leaves 22 throw one away 21

1/3rd taken away leaves 14
Hello Eargasms,

You are there, but still you can beat it.
Ok Third time lucky then ...........................