Answer from SAP forum

http://forumsa.sdn.sap.com/thread.jspa?messageID=8599950&
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A friend of mine sent me a puzzle from IBM month puzzle:

Present a computation whose result is 5, being a composition of commonly used mathematical functions and field operators (anything from simple addition to hyperbolic arc-tangent functions will do), but using only two constants, both of them 2.
It is too easy to do it using round, floor, or ceiling functions, so we do not allow them.
(http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/Challenges/January2010.html)

It seems no one has solve it yet. I would like have a try:

LOG(2)=LN(2)/LN(10)

THEN LOG(10)=LN(2)/LOG(2)

THEN EXP(LOG(10))=EXP(LN(2)/LOG(2))

THEN 10=EXP(LN(2)/LOG(2))

THEN 5=EXP(LN(2)/LOG(2))/2

THEN EXP(5)=EXP(EXP(LN(2)/LOG(2))/2)=√(EXP(LN(2)/LOG(2)))

Then the answer is 5=LN(√(EXP(LN(2)/LOG(2))))


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It seems there are lots of junior mathematicians interested in IBM puzzles. But most of those puzzles rely too much on mathematical knowledge and skills, not suitable for people are not junior mathematicians. I perfer to design less scholastic puzzles.

所有跟帖： 

• 回复：Answer from SAP forum -jinjing- ♀ (146 bytes) () 01/06/2010 postreply 20:37:10

• 还有更好的办法。用LOG2相当于又用了一个10 -康MM- ♀ (0 bytes) () 01/07/2010 postreply 06:12:25

• 回复：还有更好的办法。用LOG2相当于又用了一个10 -jinjing- ♀ (130 bytes) () 01/07/2010 postreply 07:49:01

• wrong:5=exp(log2)/2. -jinjing- ♀ (0 bytes) () 01/07/2010 postreply 13:33:46

• exp 相当于用了一个 e -WilliamLuo- ♂ (0 bytes) () 01/08/2010 postreply 10:23:48

• 回复1...1-11..=1.1=5. -jinjing- ♀ (84 bytes) () 01/09/2010 postreply 12:33:41

• 5=s(2*2),s is seccessor function -jinjing- ♀ (63 bytes) () 01/09/2010 postreply 05:42:09

• 2×2+2/2 - LN 里的e，√里的2，LOG的10 也该算吧！ -yuhaian- ♂ (91 bytes) () 02/23/2010 postreply 17:29:16

• 第二行错，应为ln10=ln2/log2 -心如海- ♀ (0 bytes) () 04/11/2010 postreply 02:50:36