| n | expression | number of functions used |
| 1 | 3SF | 2 |
| 2 | 3!SF | 3 |
| 3 | 3 | 0 |
| 4 |
3!!SSF!!SSSSSSSF!SSSSSSF =3!!SSF!!S^7F!S^6F |
23 |
| 5 | 3!!SSF | 5 |
| 6 | 3! | 1 |
| 7 |
3!!!SSSSSSSSSSSF = 3!!!S^11F |
15 |
| 8 |
3!!!SSSSSSSSSSSF!SSF = 3!!!S^11F!SSF |
19 |
| 9 |
3!!SSF!SF!SF!SSSSSSSSSSSF!SSSSSSSSSSF =3!!SSF!SF!SF!S^11F!S^10F |
36 |
| 10 | 3!!SSF!SF | 8 |
Here is Henry Shin's solution:
A few easy ones follow: floor(sqrt(3)) = 1 floor(sqrt(3!)) = 2 floor(3) = 3, or 3 = 3 floor(sqrt(sqrt((3!)!))) = 5 floor(sqrt(5!)) = 10 // from above, 5 is representible from single 3. floor(sqrt(sqrt(sqrt(10!)))) = 6 // from above, 10 is representible from 5, which in turn is representible from 3 or 3! = 6 "easy" I mean the calculations can be done with standard 4 function hand-held calculators. Now for the tricky ones, we need some form of approximation scheme for factorials. The standard trick is noting ln n! = ln 1 + ln 2 + ... + ln n as sum of areas of a rectangular figures whose base lies on x axis of xy cartesian plane and height taken as to make the total area larger or smaller than the area under the curve of y = ln x in xy plane. After a few symbol manipulations, it gives n*ln n - n + 1 < ln n! < (n+1)ln n - n + 1 Now I throw in n = (3!)! = 720 which gives 720*ln 720 - 719 < ln (720!) < (721)ln 720 - 719 since only factorial, square root, and floor function are allowed, consider (720)!^(1/2^k), or (1/2^k)*(720*ln 720 - 719) < ln (720!)^(1/2^k) < (1/2^k) * (721*ln (720) - 719) now the calculations on both lower and upper bounds for value k = 11 is "easy," giving 1.961943785... < ln (720!)^(1/2^11) < 1.96515631... or 7.113140054... < (720!)^(1/2^11) < 7.136027939... so floor(((3!)!)!^(1/2^11) ) = 7. and using the same approximation scheme for factorials and exploiting a property of logs (the useful one here being ln x^p = p*ln x), we can show that floor((10!)!^(1/2^25) ) = 4. to get 8, floor(sqrt(sqrt(7!)) = 8 Note that though the problem asks to find 8 from single 3, since 7 is derived from single 3, deriving 8 from single 7 is sufficient. The trickiest one was 9, at least I couldn't find a simpler way: Floor[Floor[Floor[Sqrt[8!]]!^(1/2^8)]!^(1/2^5)] = 9 The expression above seems un-motivated but it's actually just a doodle of getting from 8! to doing some square roots 'til we get two digit number, then factorial and square roots, etc.. In summary, floor(sqrt(3)) = 1 floor(sqrt(3!)) = 2 floor(3) = 3, or 3 = 3 floor(sqrt(sqrt((3!)!))) = 5 floor(sqrt(5!)) = 10 floor(sqrt(sqrt(sqrt(10!)))) = 6 or 3! = 6 floor(((3!)!)!^(1/2^11) ) = 7 floor(sqrt(sqrt(7!))) = 8 floor((10!)!^(1/2^25)) = 4 // that's 25 nested square roots and finally floor(floor(floor(sqrt(8!))!^(1/2^8))!^(1/2^5)) = 9