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Answer from SAP forum

Williamluo 2010-01-06 16:09:49 ( reads)

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.

跟帖(9)

jinjing

2010-01-06 20:37:10

回复:Answer from SAP forum

康MM

2010-01-07 06:12:25

还有更好的办法。用LOG2相当于又用了一个10

jinjing

2010-01-07 07:49:01

回复:还有更好的办法。用LOG2相当于又用了一个10

jinjing

2010-01-07 13:33:46

wrong:5=exp(log2)/2.

WilliamLuo

2010-01-08 10:23:48

exp 相当于用了一个 e

jinjing

2010-01-09 12:33:41

回复1...1-11..=1.1=5.

jinjing

2010-01-09 05:42:09

5=s(2*2),s is seccessor function

yuhaian

2010-02-23 17:29:16

2×2+2/2 - LN 里的e,√里的2,LOG的10 也该算吧!

心如海

2010-04-11 02:50:36

第二行错,应为ln10=ln2/log2