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 The five 3īs (Posted on 2008-07-07)
This is a real story. A long long time ago, based on the "Four fours" problem, I wondered if I could do the same using exactly five 3īs, the restrictions a bit tighter: I could use only the 4 basic math operations, exponentiation, factorial, and all parentheses I may need. Besides, I didnīt disallow me to join two "3īs" to make "33".

Using this, and only this, I succeeded in writing expressions for all integers from 0 to 100.

To narrow your work, since a great number of integers can be easily obtained, can you find expressions for 47, 50, 56, 58, 64, 70, 71, 73, 74, 76, 77, 85, 88, 94, and 95?

 See The Solution Submitted by pcbouhid Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(3): all hundred | Comment 30 of 35 |
(In reply to re(2): all hundred by Dej Mar)

> Leming, you trying to see how many you can get?

Dej Mar,

Yes (and no).  Yes I was, but after seeing your post, I realize my semi-brute-force method only allowed integers in intermediate steps.

(3!)!/(3^3) = 720/27 is not one of the possibilities under my approach.

Not sure I want to start from scratch to find out what is the smallest integer that cannot be obtained.

 Posted by Leming on 2008-07-09 09:53:15

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