Trio is Back
Tom, Dick and Harry went fishing, having agreed to share the catch equally. After a successful day, they put the catch on the boat deck to be divided up later, had a big party, and went to sleep. In the middle of the night the Tom wakes up, feeling ill, and decides to go home with his share. But he sees that the catch can't be divided evenly into three because there is one fish too many. So he throws the extra fish into the lake, takes his share, and goes home. A little later the Dick wakes up, also feeling sick, and decides to go home with his share. Not realizing that Tom has already left, he also sees that the catch can't be divided equally because of an extra fish. He throws the extra fish into the water and departs. Finally Harry wakes up, ill as well, and unaware that his friends have gone, counts the fish, finds one extra, throws it away and takes his share.
The question is, What is the smallest number of fish that the fishermen could have caught during the day?
The question is, What is the smallest number of fish that the fishermen could have caught during the day?
Labels: logic, mathemagic
posted by Rajesh Lal at
11:30 AM
30 Comments:
25,
1. Takes 8 throws 1, leaves 16
2. Takes 5 throws 1, leaves 10
3. Takes 3 throws 1, leaves 6
Add 27 and get another number that will work.. not minimum, but still works, 52, 79, 106 etc.
Harry must be left with x,
where x mod 3 =1
Dick then must have been left with
3/2*x+1
Tom must have been left with
3/2(3/2*x+1)+1
or 9/4x+5/2
All must be left with integer values
upon inspection of Dicks amount we know x must be even,
upon inspection of Toms equation we see x mod 4 =2 (to have a remaining 0.5, to add to the 5/2 to make the equation an integer amount)
x mod 3 =1 numbers, 1,4,7,10,13
x mod 4 =2 numbers, 2,6,10,14
10 is the smallest match
x=10
9/4x+5/2= 9/4*10+5/2= 25
ANSWER:
Smallest possible number of fish =25
Cam
Cam,
That's good logic, but seems kind of envolved. Could you compute the least number of fish (and the same situation) for eight guys?
It hard to believe but I got more than 1.5 million!
Tom, Dick and Harry went fishing, having agreed to share the catch equally. After a successful day, they put the catch on the boat deck to be divided up later, had a big party, and went to sleep.
In the middle of the night the Tom wakes up, feeling ill, and decides to go home with his share. But he sees that the catch can't be divided evenly into three because there is one fish too many. 25 Fish
So he throws the extra fish into the lake. 24 Fish
Takes his share, and goes home. 24 Fish, less 1/3rd leaves 16 Fish
A little later the Dick wakes up, also feeling sick, and decides to go home with his share. Not realizing that Tom has already left, he also sees that the catch can't be divided equally because of an extra fish. He throws the extra fish into the water… 15 Fish
…and departs. 15 fish less 1/3rd leaves 10 Fish
Finally Harry wakes up, ill as well, and unaware that his friends have gone, counts the fish, finds one extra, throws it away…. 9 Fish
…and takes his share. 9 fish less 1/3rd leaves 6 fish
The question is, What is the smallest number of fish that the fishermen could have caught during the day? 25 fish.
Although the real question is how come Harry, after catching approx 1/3rd of the fish, upon waking up, feeling ill doesn’t realise that his share of 3 fish is a little low?
Alchohol at parties can be a real problem for Harry, and he needs to be careful. I hope he didn’t drive home.
I was so proud of myself when I solved this on my own in under two minutes. No computation, just moving numbers around.
Yes, the answer is 25. :)
We are all proud of you anonymous, btw is that your name :)
Can anyone give any solutions for 4 or more guys? I have figured the solutions for up to eight guys. It was not to hard but the sizes of the minimum Total Fish goes up quickly.
Well, I think the generel solution for n guys is:
(n((n-1)^(n-1)-1)+1).(n/(n-1))^(n-1)+(n/(n-1))^(n-2)+(n/(n-1))^(n-3)+...+n/(n-1)+1
Please advise.
For 8, the number of fishes is 16777209. The first gets 2097151 fishes, the next 1835007, 1605631, 1404927, 1229311, 1075647, 941191 and the last one 823542.
ok, after some simplification the number of fishes is:
n^n - n + 1
where n is the number of fishermen.
Yeah, Miguel is exactly correct.
There is a problem with Miguel's formula n^n - n + 1.
i.e. it doesn't work for most n.
n=1 formula =1, should be 4
n=2 formula =4, should be 7
n=3 formula =25 WORKS !
n=4 formula = 253, should be 79
n=5 formula = 3121, should be 241
n=6 formula = 46651 should be 727
etc....
Good effort though
Cam
For 8 guys I get 6559 as the minimum
6559-1=6558, 6558*2/3=4372 #1
4372-1=4371, 4371*2/3=2914 #2
2914-1=2913, 2913*2/3=1942 #3
1942-1=1941 1941*2/3=1294 #4
1294-1=1293 1293*2/3=862 #5
862-1=861 861*2/3=574 #6
574-1=573 573*2/3=382 #7
382-1=381 381*2/3=254 #8
Slightly less than a million ;)
Cam
For n=9 I predict 19681
For n=10 I predict 59047
For n=11 I predict 177145
where each successive answer =
3*(Previous answer +1) + 1
(provided n >2, since n=1 is a strange case)
Cam
As I said before, Miguel is exactly correct when he said
Total Fish is n^n - n + 1.
I did not review his previous equations... I just check this last one.
There errors in some others above.
For example Anon said
"For 8 guys I get 6559 as the minimum".......
NO because the 6559 mod 8 is 7 not 1.
I see other errors also.
Please review my tables on my blog
http:\\Ragknot.blogspot.com
I see someone who keeps dividing by 3 even though the number of guys is more than 3.
I did that at first when I went to 4 guys, then I saw I needed "x mod 4" not "x mod 3"
Cam?
I looked back to see if Anon signed a name. I was suprised to see it was your equations that said eight guys and was still dividing by 3.
I think you got on the wrong track.
But like I said, I did too, but caught my error.
oops, looks like I had a brain fart. Please disregard all my post for n>3
Cam
Cam,
You've proved once again, the title of this Blog.
You fell prey the Trick of Mind.
Whilst many of the respondents have looked at the theoretical mathematics, try looking at the practicalities of the real world.
Assume that any number of fishermen start fishing at 6am and finish at 8pm (ignoring European Working Time Directives)and have the ability to catch 1 fish a minute (good going) they can catch a maximum of 840 fish each. Then you need to consider the size of the boat, the ability of the fishermen to accurately count some of the larger numbers quickly - it would take about 23 days to count to a million by one person,(mathforum.org/~sanders/geometry/GP09Million.html), and finally fishing the local stock dry.
I think that the problem exists in the real world for 5-6 people, but once you get up to 8 people with 1.67 million fish (Miguel Tato) being caught to be divided up in the outlined scenario..... it just ain't going to happen
I recommend that you waste no further time on this problem and retire to the bar! I have already..
the best answer is -3
oh i mean -2
I agree. Comment #3 by Cam was the best, detailed answer to this ToM.
Number one was the most correct and to the point.
Now let's go to the fish fry and have a couple of beers.
i mean the answer is minus 2
0 minus 2
-2-1=-3
-3/3=-1
-3-(-1)=-2
I think the correct answer is 25 fish because if to start with they have 25 fish and one person divides the fish evenly for 3 people, leaves with his share and throws one fish back then you have 16 fish left. then another person divides the fish evenly for 3 people, leaves with his share and throws one back leaving 10 fish. then the last person divides the fish evenly for 3 people, takes his share and leaves after throwing back one fish, leaving 6 fish
@ Neelu:
No, my name is not Anonymous. It's Joy. I just didn't feel like making an account. :D
yah i just figured it out with logic lol
Dear Author trickofmind.com !
In it something is. I thank for the information.
I want to quote your post in my blog. It can?
And you et an account on Twitter?
Post a Comment
Links to this post:
Create a Link
<< Home