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That is not true, the original question is completely well formed, just because there are some people just coming in to look all the time and then give ill responses and ill explanations:p :pQuote:
Originally posted by souldog
The Original Question Is Ill-formed, hence the question in response to the question.
Can we please get back on topic?
I am sorry, I posted the question but no one accepted to give me the answer, I now will post a short answer, that is related to the generalized Fibonacci number.
Does anyone want to make a question or I will go on with my numbers ?:)
i'm out
That one is easy.
Divide the 12 balls into 3 piles of 4 balls, call them A, B and C.
Compare A and B
- If A is lighter than B
Then either the wrong ball is lighter and in A, or heavier and in B.
Take 2 balls of A (say A1 and A2), 2 balls of B (say B1 and B2) and compare them with three balls in C to which you add a third ball from B (say B3).
*If the first pile is lighter then either A1 or A2 is wrong and light or B3 is wrong and heavy. So compare A1, B3 with C1, C2.
- If it's lighter then A1 is wrong and light
- If it's heavier then B3 is wrong and heavy
- If they are equal then A2 is wrong and light
* If the two piles are equal, then either A3 or A4 is wrong and light or B4 is wrong and heavy. Solve this similarly as above
* If the first pile is heavier, then either B1 or B2 is wrong and heavy. Compare one of them to C1 to see which one is wrong.- If A is heavier than B
Same argument as above, only reverse A and B- If A is the same as B
Then the wrong ball is somewhere in pile C.
Take C1, C2 on one side and C3, A1 on the other side.
* If the first pile is lighter then either C1 or C2 is wrong and light ot C3 is wrong and heavy. Use same as above to find solution.
* If the first pile is heavier, then either C1 or C2 is wrong and heavy or C3 is wrong and light. Same argument again
* If both are equal, then C4 is wrong. Compare it to A1 to know whether it is too light or too heavy.
Oh dear god no...Quote:
Originally posted by ih8mylife
you have 12 ball bearings
1 of them is either heavier or lighter
you have 1 balance that you can use 3 times
how do you find out which one is the odd ball (and you must know if its heavier or lighter)
nice work
uhateurlife, :D
problem is we had this one before... post something new
so there's a new one
suppose you have 3 numbers X and Y, both having equal number of digits, and N (no digit limitation)
and two other numbers X..X and Y..Y that are formed by puttin N times Xs and Ys together (simply next to each-other)
(ex X=45 Y=98 N=3 therefore X..X = 454545 and Y..Y = 989898)
which is bigger
X/Y or X..X/Y..Y
and why?
DO you mean to say positive whole numbers?
X...X/Y...Y = X/Y^N so It depends on whether X >= Y or X < Y
look at the example :pQuote:
Originally posted by me
X..X and Y..Y are formed by puttin N times Xs and Ys together (simply next to each-other)
(ex X=45 Y=98 N=3 therefore X..X = 454545 and Y..Y = 989898)
although they must be whole and for more convenience positive ;):rolleyes:
What about the example? Just because your example uses positive whole numbers does not mean it is a constraint on the problem. :)
Anyway, Did I not answer the question?:)
What... is the Airspeed Velocity of an Unladen Swallow?
FAST
no you did notQuote:
Originally posted by souldog
What about the example? Just because your example uses positive whole numbers does not mean it is a constraint on the problem. :)
Anyway, Did I not answer the question?:)
It does not depend on anything
look at my post
there are NO powers! (no X^N)