Timeline for Four ones and four zeros
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2023 at 18:10 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 27, 2023 at 17:25 | comment | added | Bernardo Recamán Santos | Rick Shepard: Very nice trick, but I mean "simple concatenation", which most seem to understand what it is about. For the record, yours is an alternative, if everything else fails. | |
| Mar 27, 2023 at 17:00 | comment | added | Rick Shepherd | @BernardoRecamánSantos 499 = 101^11/((.(0!))*(.(0!)))-0!, where each .(0!) becomes .1 by concatenation, if .x is allowed to be treated as -x currently is. Actual subtraction requires 0-x, costing an extra 0; this means -x either uses negation (aka unary minus or sign reversal) -- which is not listed in the question currently -- or concatenation). Still working on getting the question clarified. A mathematician friend told me that this type of problem is "really hard to make precise" -- and I don't disagree.... | |
| Mar 23, 2023 at 2:21 | history | bounty ended | Bernardo Recamán Santos | ||
| Mar 23, 2023 at 2:21 | vote | accept | Bernardo Recamán Santos | ||
| Mar 22, 2023 at 4:59 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 21, 2023 at 20:17 | comment | added | Weather Vane | I extended my generator to use fractional intermediate values and have two new results $ 549 = ((0! + 0! + 0!)! + (11!)! \times .1) \times (.1 + 0!) $ and $ 795 = ((\frac{0! + 0!}{.1})!! - .1) \times ((11! + 0!)!! + 0!) $ | |
| Mar 21, 2023 at 6:26 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 20, 2023 at 23:24 | comment | added | Bernardo Recamán Santos | Thomasi: Not for now! | |
| Mar 20, 2023 at 20:26 | comment | added | ThomasL | Is interpretation of the brackets as binomial coefficient allowed? If yes, we can do $462 = \binom{11}{5}$ with f.e. $462 = \binom{1100-0!}{101}$ | |
| Mar 20, 2023 at 20:12 | comment | added | Mike | Thanks for extending the list. | |
| Mar 20, 2023 at 18:55 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 20, 2023 at 17:43 | comment | added | isaacg | @Mike Updated, thanks! | |
| Mar 20, 2023 at 17:42 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 20, 2023 at 17:02 | comment | added | isaacg | @WeatherVane Thanks for the edit | |
| Mar 20, 2023 at 12:18 | history | edited | Weather Vane | CC BY-SA 4.0 |
typo correction
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| Mar 20, 2023 at 12:15 | comment | added | Weather Vane |
@BernardoRecamánSantos so it's just a typo in the post. The 462 = ((100!!)!!+0!)*(0!-1/11!) should be ((100!!)!!+0!)/(0!-1/11!) - edited.
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| Mar 20, 2023 at 11:38 | comment | added | Bernardo Recamán Santos | Weather Vane: 385/(5/6) does it, beautifully! | |
| Mar 20, 2023 at 8:44 | comment | added | Weather Vane | What am I missing? I get the 385*5/6 to be irrational. $320 \frac{5}{6}$ not 462. | |
| Mar 20, 2023 at 7:02 | comment | added | Mike | Awesome. :-). Because it cant find all cases do you want to put 500 - 1000 up as-well so we can all work together on the gaps ??? | |
| Mar 20, 2023 at 6:52 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 20, 2023 at 0:00 | comment | added | Eric Snyder | Excellent work! Though I think the tiebreaker should also include something for "number of characters used." $0!$ instead of just $1$ seems kinda silly. :) Also... man, I think I have to learn Rust at some point. | |
| Mar 19, 2023 at 10:33 | history | edited | isaacg | CC BY-SA 4.0 |
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| Mar 19, 2023 at 10:27 | history | answered | isaacg | CC BY-SA 4.0 |