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Title: cube root of 12167 Post by Noke Lieu on Dec 9th, 2008, 9:23pm Was sent this. Taken a while to figure out why it works, and I'm not exactly happy with my answer. (actually, took a moment to get it- but then a while to show that was true) It's futile (http://www.futilitycloset.com)indeed, but figuring out why it works was interesting. 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 7^3 = 343 8^3 = 512 9^3 = 729 With them you can find any two-digit cube root. For example, what's the cube root of 12,167? 1. Express the number in six digits (012167). Take the first three digits (012) and compare them to the blue cubes above. Find the largest cube that's less than your three-digit string, and write down its root. Here, 012 is between 8 and 27, so we write down 2. 2. Match the last digit of the number (7) to the last digit of a blue cube above (here, 27). Write down the root of that number (3). That's it. Put the two digits together (23) and that's your root: 233 = 12,167. This works for any perfect cube between 1,000 and 1 million. http://www.futilitycloset.com/2008/12/07/root-cause/ |
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Title: Re: cube root of 12167 Post by towr on Dec 10th, 2008, 1:00am [hide]The last digits for the first 10 cubes are unique, so the last digit of the cube betrays the last digit of the cuberoot. For the first digit of the cuberoot, check whether the number is between 1000 times one cube and the next. It should work for every power N where the last digits are unique for the first 10. And of course remember to discard the last N digits to find the first digit of the N-root.[/hide] |
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Title: Re: cube root of 12167 Post by Noke Lieu on Dec 10th, 2008, 2:50pm okay... that's not dissimilar to what I first did. And the part that took a while was trying to express that more formally- hence I'm not entirely happy with my solution. I started with [hide](10a+b)3[/hide] which turns to [hide]1000a3 + 300a2b +30ab2 + b3[/hide] That demonstrates why the units digit of the cube gives the units digit of the root (every other term is multiplied by at least 10...) but to explain the first step was... harder... and that's the bit that left me grasping. By examining only the first three digits (including 0) of the cube... multiplying the the "blue" cubes by 1000 to allow comparison seemed sensible. Then I lost faith in the [hide](10a+b)3[/hide] tack, and went with [hide](11a+x)3[/hide] where a+x=b yielding [hide] 1331a3 + 363 a2x + 33ax2 + x3[/hide] (yesterday, it made sense doing that- wish I'd annotated my notes... :-/ ) Had to do with determining which term gave the leading digit of the cube... |
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Title: Re: cube root of 12167 Post by towr on Dec 10th, 2008, 3:13pm For extra clarity, [hide]given digits a and b, (10 a)3 <= (10 a + b)3 < (10 a + 10)3 = (10 [a+1] )3 So you can determine a from this.[/hide] |
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Title: Re: cube root of 12167 Post by codpro880 on Dec 24th, 2008, 11:55pm That's pretty cool. |
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