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If one and one
and two and three
and five is next
of course you'll see
that two and five
and seven, twelve
will futher show
that should you delve
into the next
past one-five-six
will lead to x
and x plus which?

2006-07-06 12:41:34 · 5 answers · asked by Scott R 6 in Science & Mathematics Mathematics

Mathematicians generally shun this type of question. There's no glory in solving it, and are afraid to hazard a guess when they dont know lest they're wrong.

2006-07-06 13:01:53 · update #1

someone's getting very close...

2006-07-06 13:15:56 · update #2

whichx = (which)(x) maybe?

2006-07-06 13:25:49 · update #3

oh well, this was just an experiment in (english) linguistics. a little lengthy to go into here. everything was exact in its phrasing and typography.

2006-07-06 13:27:44 · update #4

my whole point here was that line 11, the word "will" is not exactly grammatically accurate, or at least misleading. It should be it'll or that'll (even given poetic license). What was the original question?

2006-07-06 13:33:41 · update #5

5 answers

This is obviously about Fibonacci sequences. 1 and 5 go to 6, 5 and 6 go to 11 (x) and x +6 go to the last.

So the answer should be
x=11 and which=6. I think.

Maybe by "plus" you don't mean the mathematical operation, but "and." In that case, x=11 and which=x+6=17

So the question is at the beginning (as Yahoo designed), and not in the details.

Since which=17 and x=11, whichx=11•17=187

I will admit that I'm not very good at these.

The original question is "What is x, and whichx?"

2006-07-06 13:09:48 · answer #1 · answered by Eulercrosser 4 · 1 1

The pattern is
1+1=2
1+2=3
2+3=5

2+5=7
5+7=12

1(next past)+(5+ 6)=12

Will lead to x and x plus 1

2006-07-06 20:52:23 · answer #2 · answered by nedoglover 4 · 0 0

x + 156

2006-07-06 19:55:58 · answer #3 · answered by georgephysics13 3 · 0 0

I think Eulercrosser is on the right track, if not already correct.

2006-07-06 20:11:37 · answer #4 · answered by arbeit 4 · 0 0

X +!X

2006-07-06 19:49:25 · answer #5 · answered by Phillip R 4 · 0 0

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