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Complete the sequence
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Original Post
I made this up in lecture today - there are two versions to this. I'm posting the harder version with a clue that should remove any dualities or multiple solutions. 1 1 1 2 _ _ 3 _ _ The clue: 1 < 1 < 1 IMPORTANT NOTE: Admittedly, this sequence is obscure, but should only have one set of congruent answers. Post the minimal possible solution from that set (as in: of two members of one set, for instance 3/3 and 6/6, 6/6 is smaller than 3/3, although not minimal in the set in this case).
Normally, sequence puzzles are fairly well defined, such that a valid "wrong" answer is impossible. I would hazard a guess that Override's answer is "wrong".
This is normally a symptom of odd-one-out puzzles, which I hate for that very reason, and don't understand why they are found on so many IQ tests.
This is normally a symptom of odd-one-out puzzles, which I hate for that very reason, and don't understand why they are found on so many IQ tests.
Quote:
Admittedly, this sequence is obscure, but should only have one set of congruent answers. Post the minimal possible solution from that set (as in: of two members of one set, for instance 3/3 and 6/6, 6/6 is smaller than 3/3, although not minimal in the set in this case).
I don't understand how 6/6 is "smaller" than 3/3 when both represent the same value, and especially since 6 > 3.
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Original post by kSquared Quote:
Admittedly, this sequence is obscure, but should only have one set of congruent answers. Post the minimal possible solution from that set (as in: of two members of one set, for instance 3/3 and 6/6, 6/6 is smaller than 3/3, although not minimal in the set in this case).
I don't understand how 6/6 is "smaller" than 3/3 when both represent the same value, and especially since 6 > 3.
While it's true that 6 > 3, you need to pay attention to the denominators. ie. 1/6 is smaller than 1/3. I don't understand the number sequence above however.
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Original post by Crispy
I made this up in lecture today - there are two versions to this. I'm posting the harder version with a clue that should remove any dualities or multiple solutions.
1 1 1 2 _ _ 3 _ _
The clue:
1 < 1 < 1
Your clue makes no sense at all. But I will take a shot :
1 1 1 2 2 4 3 3 9 (1 * 1 = 1 => 2 * 2 = 4 => 3 * 3 = 9)
You should post more of the sequence, because at the moment there are thousands of possible correct answers and it's just a guessing game.
A better clue might help too. One is smaller than one is smaller than one does not make sense.
My 'guess' is:
1 1 1 2 2 1 3 3 1
A better clue might help too. One is smaller than one is smaller than one does not make sense.
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Your clue makes no sense at all. But I will take a shot :
1 1 1 2 2 4 3 3 9 (1 * 1 = 1 => 2 * 2 = 4 => 3 * 3 = 9)
My 'guess' is:
1 1 1 2 2 1 3 3 1
1 1 1 2 _ _ 3 _ _
1*1=1 2*1=2 3*1=3 ==>
1 1 1 2 1 2 3 1 3
?????
1*1=1 2*1=2 3*1=3 ==>
1 1 1 2 1 2 3 1 3
?????
This whole exercise appears utterly pointless. When someone gets the answer, this had better be good, and I doubt it will, since whenever one of these has happened before the answer was just made hopelessly obscure by how ill-defined the problem was.
Oh, and nobody's done 1 1 1 2 1 1 3 1 1 yet, or 1 1 1 2 4 8 3 9 27.
Oh, and nobody's done 1 1 1 2 1 1 3 1 1 yet, or 1 1 1 2 4 8 3 9 27.
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Original post by Crispy Quote:
Original post by dorix
I was thinking it could be:
1 1 1 2 4 8 3 9 27
(1**1 1**2 1**3 2**1 2**2 2**3 3**1 3**2 3**3)
Your solution is correct. And should be the only correct one if you read my original post word-for-word and pay attention to every detail.
Cool, but I still don't understand how the original clues eliminate any of the other choices given -- or suggest this one over any others. I pretty much just pulled that out of a hat.
So how can 1 < 1 < 1 ever be true?
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Original post by dorix
So how can 1 < 1 < 1 ever be true?
Do not view the 1's as numbers, but rather as different instances of members from the same set.
Let's say the relationship between two members of a set is: nj < nj+1, where njk = njk.
In human terms this means that x < y if x = nk and y = nk+m, where m > 0.
Now imagine if n = 1.
Finally, do not think that 1 < 1, but think that the expression behind one 1 is less than the expression behind the other 1: 11 < 12 < 13.
This isn't a problem from the realm of mathematics, so don't treat it as one, but rather a problem from the real of set theory. Minimization is a set theory function that finds the smallest possible solution to a problem - ergo, for instance, given that the set is held together by the relation that q < r, and that the set only contains letters, the minimized expression of the relation is b < a, not q < r.
Does that make any sense (read: did I explain that right because I could be wrong in my explanation?)?
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