Nov 22, 2009
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Logical Question

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Read Latest Entries..: (Post #11) by iGuest on Sep 5 2009, 06:23 PM.
Replying to csp4.0 Hi friends I have here a simple solution The first line have 1 (one time 1) so next 11 (First write how many times a digit occurs than write that digit) Mean in second line (11) 1 is two(2) times so first 2 than 1... 21 1211 .. .. ... . 1113213211 and next will be 31131211131221 and more next will be 13211311123113112211 -reply by satyajeet vyas...
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Logical Question

csp4.0
I knew of this question and its answer so i wanted to share it with you people here on Trap17 itself... Anyway this question is designed to to test your logic, not your mathematic skills despite it being numbers
ok, the question:

1
11
21
1211
111221
312211
13112221
1113213211
*find this line*

I'm sorry if this is posted in the wrong section but I couldn't manage to find a 'logic questions' section.
Anyway, Feel free to discuss this, but please keep the answer quiet, if you happen to find the answer please PM me and I will tell you if it is correct.

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strange-garden
QUOTE(csp4.0 @ Jan 14 2008, 11:23 PM) *
I knew of this question and its answer so i wanted to share it with you people here on Trap17 itself... Anyway this question is designed to to test your logic, not your mathematic skills despite it being numbers
ok, the question:

1
11
21
1211
111221
312211
13112221
1113213211
*find this line*

I'm sorry if this is posted in the wrong section but I couldn't manage to find a 'logic questions' section.
Anyway, Feel free to discuss this, but please keep the answer quiet, if you happen to find the answer please PM me and I will tell you if it is correct.


Easy.

three ones, one three, one two, one one, one three, one two, two ones


31131211131221

amiright? amiright?

and just to show you that I'm not cheatng...The next answer is

one three, two one, one three, one one, one two, three one, one three, one one, two two, one one

13211311123113112211

 

 

 


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budsu
well this must be 10 nuber line

i think it starts with 4

4132132111

am i right ?

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csp4.0
QUOTE(strange-garden @ Jan 15 2008, 10:51 AM) *
Easy.

three ones, one three, one two, one one, one three, one two, two ones
31131211131221

amiright? amiright?

and just to show you that I'm not cheatng...The next answer is

one three, two one, one three, one one, one two, three one, one three, one one, two two, one one

13211311123113112211


Yes, you are right.

QUOTE(budsu @ Jan 15 2008, 08:00 PM) *
well this must be 10 nuber line

i think it starts with 4

4132132111

am i right ?


You, my friend are not right. please refer to the person above (2nd post) for the answer.

This question has never been hard to many people, it is just how long it took them to find the answer. If I didn't tell you that it was logical many might have though t it was mathematical...

the correct answer is : 31131211131221

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budsu
i doesnt make sense tongue.gif

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strange-garden
YAYYY!! I winnnn!! Girl powerrrr!!

and it does makes sense. You're describing the number above it, for example

1122321

You work it out by saying, two one (because the first is two ones) and then two two (because there's two twos) and the one three (there's only one three next).

So it ends up like this:

two one two two one three one two one one

which translantes to:

2122131211


Does it make sense now?

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Tourist
Very interesting!!!

Before I saw the explanation in the last post, I did not understand the logic. I was trying to solve this in mathematical way, but finding no solution lol.

I have already added this question in my own forum and waiting for the reply of forum members of that site. I think, they will also enjoy this Logical Question.

Thank you for this interesting Logical Question.


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tricky77puzzle
I've also noted something:

(1) 1 = 1
(11) 1 + 1 = 2
(21) 2 + 1 = 3
(1211) 1 + 2 + 1 + 1 = 5
(111221) 1 + 1 + 1 + 2 + 2 + 1 = 8

It's too bad 312211 breaks the sequence (10), because this would have been pretty cool as the Fibonacci sequence. Actually, if we disallow 3's and say that 111 must be written as 2111 instead, the next sequence number would be 21112211, which adds up to 11. Still missing a few. If 22 must be written as 1212, then we come up with 2111121211, which is 13, but that's kind of pushing the envelope. After that would be 12212112111221, which is 20, which is still not enough.

Anyway, here's something to do: try to prove that if the sequence starts with 1, it is impossible for the digit 4 to appear in this sequence.

EDIT: If we say that "1111....1" must be written as 21 and then one "11" for every trailing 1, we get 1221111112111221, which is equal to 21. After that would be 1112122111111111122111121211, which is 35. It just doesn't work, for some reason.

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FeedBacker
Replying to budsu


NO!

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iGuest-top
You just read the first line and write your sound in the second line.
1 is one 1 ; you write it down
11 is two 1 ; write it down
21 is one 2 and one 1; write it down
1211

Clear?

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Latest Entries

iGuest
Replying to csp4.0 Hi friends I have here a simple solution The first line have 1 (one time 1) so next 11 (First write how many times a digit occurs than write that digit) Mean in second line (11) 1 is two(2) times so first 2 than 1... 21 1211 .. .. ... . 1113213211 and next will be 31131211131221 and more next will be 13211311123113112211 -reply by satyajeet vyas

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