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04-24-2018 , 08:08 PM
Quote:
Originally Posted by eyebooger
Spoiler:
Trial 1 through 10, all triplets.
Assume "close" for 111, 333, 999.

Trial 11: 144
If "close", you know digit 1 is in the 1st spot.

If "fail",
Trial 12: 414
If "close", digit 1 is in the 2nd spot.
If "fail", digit 1 is in the 3rd spot.

Trial 13: 344
If "close", digit 3 is in 1st spot, solved.
If "fail", digit 9 is in 1st spot, solved.
Spoiler:
What about if 111 and 999 are close, the rest fails?
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04-24-2018 , 08:10 PM
Quote:
Originally Posted by iraisetoomuch
I'm pretty sure you're both +1 on the low end and maybe on the high end

Spoiler:
because If you get 9 with no close you obv know there is a 0 so it could take only 9.

I don't think this applies to the high end but I'm not sure.
True
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04-24-2018 , 08:11 PM
irtm is correct about min, I missed that scenario.
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04-24-2018 , 08:23 PM
Quote:
Originally Posted by e_holle
Spoiler:
What about if 111 and 999 are close, the rest fails?
Spoiler:
144 then 414.
If both fail, solved (991)
If both close, solved (119)

If 144 close 414 fail, do 441 to solve.

Still max 13.
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04-24-2018 , 08:35 PM
Quote:
Originally Posted by eyebooger
Spoiler:
144 then 414.
If both fail, solved (991)
If both close, solved (119)

If 144 close 414 fail, do 441 to solve.

Still max 13.
Yeah, I saw that after. I'm blaming it on the fever.
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04-25-2018 , 01:42 AM
eyebooger is correct
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04-25-2018 , 08:30 AM
In the ideal scenario, what is the minimum amount of tries before you can be 100% the answer is correct?

Spoiler:
I'm thinking 9. Try 111 - 999, FAIL on all, you know 000 is correct.
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04-25-2018 , 08:47 AM
Quote:
Originally Posted by housenuts
In the ideal scenario, what is the minimum amount of tries before you can be 100% the answer is correct?
Spoiler:
111 - close
222 - close
333 - close
144 - close
424 - close or fail, solved
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04-26-2018 , 12:01 AM
Absolute ideal?

Spoiler:
144 - close
424 - close
443 - close
444 - fail
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04-26-2018 , 02:36 AM
yeah I'd agree with both those

ibavly has the quickest possible solution and eyebooger has the quickest possible solution using the optimal strategy
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04-26-2018 , 02:46 AM
Easy problem but one I liked

A and B represent digits

A/B=A6/6B=A66/66B=A666/666B=A6666/6666B

What are the possible A and B combos?
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04-26-2018 , 04:27 AM
Spoiler:
14 and 66
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04-26-2018 , 06:37 AM
Spoiler:
And 25?
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04-26-2018 , 08:57 AM
Is there an easy way to solve that besides trial and error?
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04-26-2018 , 09:50 AM
It is simple to show that B < 7. 6 is obvious, then the other options are easy to check. I think it is also not too hard to show that exactly one of them must be even when not both 6.
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04-27-2018 , 10:44 AM
this weeks riddler

Quote:
What is the fastest way to fill up a text editor with a string of 1 million of the same character? (Let’s go with the letter “i”.)

There are a lot of variables here. You can type “i’s” at a certain rate, maybe around five per second, by simply pressing its key repeatedly. You can also hold down the key, initially getting a single “i,” and then after a “repeat delay” of about half a second, getting a quickly repeating stream of “i’s” at a “repeat rate” of about 30 per second. You can also use copy and paste. If you release the “i” key, you can hit Ctrl+A then Ctrl+C, then hit the right arrow key, and finally Ctrl+V, selecting all your text, copying it and pasting it to what you had already. (Replace Ctrl with Command if on a Mac, of course.) This process costs you about a second from “i” key release to initial depress of Ctrl+V. If you hold down Ctrl+V, there is the same repeat delay and repeat rate that then generates a bunch of copies of your clipboard very quickly.

So the questions are: How big should you make the original edition of your clipboard before you transition to the more efficient copy/paste? Then, how long should you stick with that clipboard before going back to the Ctrl+A and growing your clipboard again?
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04-27-2018 , 10:46 AM
That is brutal to read as text. **** need to be in a table.
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04-27-2018 , 12:01 PM
If we're assuming that our only parameters are how many original 'i's we have, and how often we repeat the ctrl+a, the best I get it starting with 12 'i's and hitting ctrl+a every 16 pastes.

12 'i's = .866s
paste 16 times = 2s

need to repeat 4 times - 12*17^4 = 1002252

4*2+.866 = 8.866s

Sort of surprised that it can get this fast. Not at all optimized though, I guess you could shave some milliseconds if you're willing to use an inconsistent number of pastes each time.
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04-27-2018 , 12:25 PM
Pasting 16 times takes 16/30 seconds not 15/30 seconds. So you're either pasting 15 times for a size of 16 in 2s or 16 times for a size of 17 in 2.03s I think.

Best I can get is 9.008s with an initial paste of 'I'+ 15x 'I' and then repeat every 15.

I dunno.
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04-27-2018 , 12:57 PM
it's 15/30 because you click right for one copy (included in the 1 second) and then the first ctrl+v is instantaneous, so after the .5s delay you get another 15 copies in .5s for 17 total copies.
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04-27-2018 , 01:01 PM
Ah yeah. That makes sense.
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05-17-2018 , 11:55 PM
0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

You can use any mathematical symbols in the spaces provided to make all above algebraic expressions true.
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05-18-2018 , 02:04 AM
Spoiler:
(0! + 0! + 0!)!
(1+1+1)!
2+2+2
3x3-3
Root4 + Root4 + Root4
6+6-6
7-7/7
8-Root(Root(8+8))
Root9 x Root9 - Root9


Or is that considered cheating?
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05-18-2018 , 02:11 AM
Spoiler:
Seen this problem before and pretty sure I remember that being essential. Especially !
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05-18-2018 , 04:14 AM
I never believed 0!=1, no matter what they told me...
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