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| 13 Sep 2015 01:37 PM |
Lets take the number 27, we want to find a root for this number. So we first take the number 2
sqrt(27) = ucky decimals
but lets take the third root of it thirdroot(27) = 3
Wallah! We found a number that leaves a nice number behind, but the problem is we guessed and checked. How do we find out what number we can use without guessing and checking? |
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| 13 Sep 2015 01:39 PM |
| factor 27? into 3*9 then factor it into 3x3x3? Idk, factor trees may work |
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| 13 Sep 2015 02:00 PM |
| Im talking about doing it with a computer |
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cntkillme
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| Joined: 07 Apr 2008 |
| Total Posts: 44956 |
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| 13 Sep 2015 02:02 PM |
But the thing is, there are practically an infinite number of 'roots' you can do.
the 3.14th root of 36.337838880174696063223495599519 is 2 |
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| 13 Sep 2015 02:09 PM |
| yes do a factor tree. then only accept integers as potential "roots" |
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| 13 Sep 2015 02:10 PM |
| I could do decimals too, even in the final result, I just dont want it to be gigantic like 5 decimals on any three numbers max |
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| 13 Sep 2015 02:16 PM |
I came up with this for guessing and checking. I dont know how I would code a factor tree
local code = 32542390589034985689034859034985890234859034895034859043 for i = 1,1000,.00001 do local num = (code ^(1/i)) if num% 1 ~= 0 then print('FOUND',i,num) break end wait() end |
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| 13 Sep 2015 02:34 PM |
| My last resort would be to find a web api then do get requests lol |
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| 13 Sep 2015 02:37 PM |
"dont know how I would code a factor tree" If you even had the processing power to factor a number that big xD |
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litalela
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| Joined: 30 Mar 2010 |
| Total Posts: 6267 |
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| 13 Sep 2015 02:40 PM |
well i mean
sqrt(27)
find perfect square factors
if they are there then
take out perfect square, square root it
end result
3*sqrt(3)
basic math
➳Lɪᴛᴀʟᴇʟᴀ ɪs ᴍʏ ɴᴀᴍᴇ, Lᴜᴀ ɪs ᴍʏ ɢᴀᴍᴇツ |
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