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Title: All the digits without repeats Post by John C. on Apr 9th, 2005, 9:52am Alright guys =) i need an answer to this question and its been bugging me... if you manage to figure it out PLEASE email me at shadowmage@comcast.net!!! thank you soo much *start problem* : Find a number that uses all the whole digits between 0 and 10, uses each number only once, and is evenly divisable by all whole numbers between 2 and 18. Ex. 1,234,567,890 is a valid number that uses each number only once, but its not the correct answer because it is not evenly divisable by the number 17. Good luck. |
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Title: Re: Just Looking for an answer Post by Sjoerd Job Postmus on Apr 9th, 2005, 10:59am EDIT: Ignore this post, it contains an error. Thanks go to T&B for telling me. on 04/09/05 at 09:52:57, John C. wrote:
Find the smallest number that's divisable by all numbers from 1 to 18. ( div by 0 won't work ;) ) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18 Don't have to worry 'bout 9, 6, 3, 2 and 1 anymore. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 17 18*17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 16 18*17*16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 But, 18*16 is divisable by 12 ( 18*2 is, so 18*2*8 is ) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 15 18*17*16*15 From which follows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 14 18*17*16*15*14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 13 18*17*16*15*14*13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up: 11 ( 12 is already fixed ;) ) 18*17*16*15*14*13*11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Next up? Done! So, let's get our calculator! Multiplying gets me to [hide]13366080[/hide]. So, we need to find a number using these digits only once, that's divisable by this number. Notice how we are 2 tokens short? So, we could get out our calculators, and tactically multiply this string with 75 up to 750, and checking if it matches ;)... |
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Title: Re: Just Looking for an answer Post by THUDandBLUNDER on Apr 9th, 2005, 11:16am SJB, your number is not divisible by 11. I get the smallest such number to be [hide]25*32*5*7*11*13*17 = 24504480[/hide] But this is from 51 to 403 times too small. At the moment I can't think of a good way to find out exactly how many times too small. |
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Title: Re: Just Looking for an answer Post by Sjoerd Job Postmus on Apr 9th, 2005, 11:52am on 04/09/05 at 11:16:05, THUDandBLUNDER wrote:
Ok, I've found out an error in my algorithm, but your answer is clouded as well. By a factor two, that is. (That's what you get for abbr. my name wrongly ;) ) [hide]24*32*5*7*11*13*17 = 12252240[/hide] An algorith to find the smallest number(i) possible that's dividable by all numbers up to n i = 1; for ( x=2; x <= n; x++ ) if isprime(x) { y = x; while ( y < n/x ) { y *= x; } i *= y; } In other words. For every prime ranging from 2 to n, find the highest power possible that's smaller or the same as n. Multiply those For this situation, 18... 24 * 32... close to t&b's answer. My first 'guess' was wrong to: I skipped multiplying by 11. EDIT: Fixing some bugs ;) |
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Title: Re: Just Looking for an answer Post by THUDandBLUNDER on Apr 9th, 2005, 1:13pm Sjoerd Job Postumus :P, your number is not evenly divisible by 16. |
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Title: Re: Post by Sjoerd Job Postmus on Apr 9th, 2005, 1:18pm on , N/A wrote: 32? Correct me if I'm wrong... but isn't 32 out of the 2 - 18 range? ;)... 12252240 / 2 = 6126120 12252240 / 3 = 4084080 12252240 / 4 = 3063060 12252240 / 5 = 2450448 12252240 / 6 = 2042040 12252240 / 7 = 1750320 12252240 / 8 = 1531530 12252240 / 9 = 1361360 12252240 / 10 = 1225224 12252240 / 11 = 1113840 12252240 / 12 = 1021020 12252240 / 13 = 942480 12252240 / 14 = 815160 12252240 / 15 = 816816 12252240 / 16 = 765765 12252240 / 17 = 720720 12252240 / 18 = 680680 --- All conditions met --- Now for the final answer? Still calculating ;)... trying to get a program to do it, because I'm lazy :) EDIT: That's what I get for jumping on the gun... but still, my answer is div'able by 16 ;) |
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Title: Re: Just Looking for an answer Post by Grimbal on Apr 9th, 2005, 4:12pm I just tested all multiples of 16*9*5*7*11*13*17 for unicity of digits and for divisibility, it gives: [hide]2438195760 3785942160 4753869120 4876391520 [/hide] |
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Title: Re: Just Looking for an answer Post by John C. on Apr 9th, 2005, 9:37pm wow, you guys/gals are really on top of things thanks soo much! ;D ;D ;D |
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Title: Re: Just Looking for an answer Post by THUDandBLUNDER on Apr 9th, 2005, 10:40pm on 04/09/05 at 21:37:59, John C. wrote:
I would have preferred a bit more analysis. Actually, the first two answers are not evenly divisible by 16. |
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Title: Re: Just Looking for an answer Post by towr on Apr 10th, 2005, 6:59am on 04/09/05 at 22:40:19, THUDandBLUNDER wrote:
And they'd better be considering they're multiples of 16*9*5*7*11*13*17 |
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Title: Re: Just Looking for an answer Post by John C. on Apr 10th, 2005, 8:40am True, BUT, even if it is not correct the first time, i did not expect such a snappy response =) so thank you for atleast getting back to it quickly! ;) |
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Title: Re: Just Looking for an answer Post by THUDandBLUNDER on Apr 10th, 2005, 10:26am on 04/10/05 at 06:59:54, towr wrote:
The number must be divisible by 32. evenly =/= even ::) |
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Title: Re: Just Looking for an answer Post by towr on Apr 10th, 2005, 10:29am ?!? Why must it be divisible by 32? |
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Title: Re: Just Looking for an answer Post by THUDandBLUNDER on Apr 10th, 2005, 10:55am on 04/10/05 at 10:29:01, towr wrote:
Because the problem statement says 'and is evenly divisable by all whole numbers between 2 and 18', and I interpret that to mean that if you divide the number by another that is between 2 and 18 then you will get an even number. But maybe that is not what was intended? Or perhaps my interpretation is wrong? If, as may be the case, 'evenly divisable' means simply 'divisible' then the word "evenly" would seem to be completely superfluous. |
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Title: Re: Just Looking for an answer Post by towr on Apr 10th, 2005, 12:08pm I do think 'evenly' was superfluous. And just a way to say the parts were the number is divided into are all equal. Well, we only needed one answer, so given that at least the latter two fit under either interpretation that should suffice :P |
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Title: Re: Just Looking for an answer Post by Sjoerd Job Postmus on Apr 11th, 2005, 10:07am Something on 'evenly divisible': http://www.testmagic.com/forums/showthread.php?t=22143 A is evenly divisible by B if A MOD B = 0. |
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Title: Re: Just Looking for an answer Post by towr on Apr 12th, 2005, 7:27am on 04/11/05 at 10:07:33, Sjoerd Job Postmus wrote:
And A is divisible (http://mathworld.wolfram.com/Divisible.html) by B, under the same condition. So 'evenly' doesn't add anything (except confusion). |
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Title: Re: Just Looking for an answer Post by John C. on Apr 14th, 2005, 9:56am :-/ well sorry for the confusion folks...thats exactly how i got the question so thats exactly how i posed it ... but now we all know a little bit more about the term "Divisible" |
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Title: Re: Just Looking for an answer Post by asterex on Apr 14th, 2005, 10:11am If evenly divisible means you end up with an even number, have you ever heard anyone use the term "oddly divisible"? The reason for saying evenly is because technically any number is divisible, i.e. able to be divided, by any number. One hundred is divisible by three. But if you have one hundred gold coins, they can not be evenly divided among three people without chopping a coin into pieces. |
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Title: Re: Just Looking for an answer Post by Grimbal on Apr 15th, 2005, 1:02am I would even say that "even" is a shortcut for "evenly divisible in 2 parts", where evenly doesn't mean that there is no rest, but that the parts are equal. |
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Title: Re: Just Looking for an answer Post by Nick S on Nov 20th, 2005, 1:57pm Well, the problem with all your answers is that they either use a number too many times or they dont use the other ones enough. They must use 0-9 and only use each once. |
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Title: Re: Just Looking for an answer Post by towr on Nov 20th, 2005, 2:58pm on 11/20/05 at 13:57:59, Nick S wrote:
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Title: Re: Just Looking for an answer Post by Sjoerd Job Postmus on Nov 21st, 2005, 1:24pm Funny thing to notice is that none of the numbers are divisable by 19 ;)... Just checked, out of boredom, if we could add 19 to it too, to decrease the list down to 1 item, or less then the 4 we have. But, none of them are div. by 19, so it doesn' work out |
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Title: Re: Just Looking for an answer Post by Grimbal on Nov 22nd, 2005, 5:27am Maybe that's why the problem statement asks for 2-18? |
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Title: Re: Just Looking for an answer Post by Sjoerd Job Postmus on Nov 22nd, 2005, 10:11am on 11/22/05 at 05:27:45, Grimbal wrote:
True, good point. But that doesn't make me not explore new possibilities. |
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