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Phi Proximity Poser (Posted on 2023-02-07) Difficulty: 3 of 5
Using each of the digits 1 to 9 exactly once, and the standard arithmetic operators, +,-, *, /, and, () as many times as required; come up with a value that is closest to φ.
This means, minimizing abs(V - φ), where V is the obtained value and, abs is the absolute function.

(I) Use only the individual digits. Concatenations like (98-34)/(1*2) - (56/7) are NOT allowed.

(II) Concatenations are allowed.

Note: φ is the golden ratio or the golden mean.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
a start for part (II) | Comment 1 of 3
One strategy for Part II:
Find the largest two adjacent Fibonacci numbers, with only digits 1 to 9 and only once each;  φ(9) divided by φ(8):
34/21 = 1.619047619047619  err=0.0010136302977241662
This is too large by about 1/1000, so let's make a fraction out of the remaining digits as close to that as possible.  
After a few attempts:  
34/21 - 8/7965 = 1.61804322482289  err=9.23607299085916E-06
34/21 - 8/7956 = 1.61804208863032  err=8.09988042904664E-06
34/21 - 7/6958 = 1.61804158283032  err=7.5940804202812E-06
  Posted by Larry on 2023-02-07 12:08:15
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