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Topic: All the digits without repeats (Read 2149 times) |
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John C.
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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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| « Last Edit: Nov 22nd, 2005, 5:39pm by Icarus » |
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Sjoerd Job Postmus
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Re: Just Looking for an answer
« Reply #1 on: Apr 9th, 2005, 10:59am » |
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EDIT: Ignore this post, it contains an error. Thanks go to T&B for telling me.
on Apr 9th, 2005, 9:52am, John C. wrote: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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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 13366080.
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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| « Last Edit: Apr 9th, 2005, 11:56am by Sjoerd Job Postmus » |
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ThudnBlunder
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Re: Just Looking for an answer
« Reply #2 on: Apr 9th, 2005, 11:16am » |
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SJB, your number is not divisible by 11.
I get the smallest such number to be 25*32*5*7*11*13*17 = 24504480
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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| « Last Edit: Apr 9th, 2005, 11:25am by ThudnBlunder » |
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Sjoerd Job Postmus
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Re: Just Looking for an answer
« Reply #3 on: Apr 9th, 2005, 11:52am » |
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on Apr 9th, 2005, 11:16am, THUDandBLUNDER wrote:SJB, your number is not divisible by 11.
I get the smallest such number to be 25*32*5*7*11*13*17 = 24504480
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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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 )
24*32*5*7*11*13*17 = 12252240
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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| « Last Edit: Apr 9th, 2005, 11:55am by Sjoerd Job Postmus » |
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ThudnBlunder
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Re: Just Looking for an answer
« Reply #4 on: Apr 9th, 2005, 1:13pm » |
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Sjoerd Job Postumus  , your number is not evenly divisible by 16.
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| « Last Edit: Apr 9th, 2005, 1:17pm by ThudnBlunder » |
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Sjoerd Job Postmus
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Re:
« Reply #5 on: Apr 9th, 2005, 1:18pm » |
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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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| « Last Edit: Apr 9th, 2005, 1:19pm by Sjoerd Job Postmus » |
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Grimbal
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Re: Just Looking for an answer
« Reply #6 on: Apr 9th, 2005, 4:12pm » |
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I just tested all multiples of 16*9*5*7*11*13*17 for unicity of digits and for divisibility, it gives:
2438195760
3785942160
4753869120
4876391520
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| « Last Edit: Apr 9th, 2005, 4:13pm by Grimbal » |
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John C.
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Re: Just Looking for an answer
« Reply #7 on: Apr 9th, 2005, 9:37pm » |
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wow, you guys/gals are really on top of things thanks soo much! 
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ThudnBlunder
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Re: Just Looking for an answer
« Reply #8 on: Apr 9th, 2005, 10:40pm » |
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on Apr 9th, 2005, 9:37pm, John C. wrote:| wow, you guys/gals are really on top of things thanks soo much! |
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I would have preferred a bit more analysis.
Actually, the first two answers are not evenly divisible by 16.
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towr
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Re: Just Looking for an answer
« Reply #9 on: Apr 10th, 2005, 6:59am » |
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on Apr 9th, 2005, 10:40pm, THUDandBLUNDER wrote:I would have preferred a bit more analysis.
Actually, the first two answers are not evenly divisible by 16. |
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Yes they are..
And they'd better be considering they're multiples of 16*9*5*7*11*13*17
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John C.
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Re: Just Looking for an answer
« Reply #10 on: Apr 10th, 2005, 8:40am » |
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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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ThudnBlunder
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Re: Just Looking for an answer
« Reply #11 on: Apr 10th, 2005, 10:26am » |
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on Apr 10th, 2005, 6:59am, towr wrote:
Yes they are..
And they'd better be considering they're multiples of 16*9*5*7*11*13*17
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The number must be divisible by 32.
evenly =/= even 
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| « Last Edit: Apr 10th, 2005, 10:27am by ThudnBlunder » |
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ThudnBlunder
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Re: Just Looking for an answer
« Reply #13 on: Apr 10th, 2005, 10:55am » |
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on Apr 10th, 2005, 10:29am, towr wrote:
Why must it be divisible by 32? |
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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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| « Last Edit: Apr 11th, 2005, 11:53am by ThudnBlunder » |
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towr
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Re: Just Looking for an answer
« Reply #14 on: Apr 10th, 2005, 12:08pm » |
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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
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towr
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Re: Just Looking for an answer
« Reply #16 on: Apr 12th, 2005, 7:27am » |
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on Apr 11th, 2005, 10:07am, Sjoerd Job Postmus wrote:
And A is divisible by B, under the same condition. So 'evenly' doesn't add anything (except confusion).
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John C.
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Re: Just Looking for an answer
« Reply #17 on: Apr 14th, 2005, 9:56am » |
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 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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asterex
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Re: Just Looking for an answer
« Reply #18 on: Apr 14th, 2005, 10:11am » |
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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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Grimbal
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Re: Just Looking for an answer
« Reply #19 on: Apr 15th, 2005, 1:02am » |
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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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Nick S
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Re: Just Looking for an answer
« Reply #20 on: Nov 20th, 2005, 1:57pm » |
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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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towr
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Re: Just Looking for an answer
« Reply #21 on: Nov 20th, 2005, 2:58pm » |
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on Nov 20th, 2005, 1:57pm, Nick S wrote:| 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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Where do you see any supposed answer that's wrong? Because the ones I see do have each digit exactly once.
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Sjoerd Job Postmus
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Re: Just Looking for an answer
« Reply #22 on: Nov 21st, 2005, 1:24pm » |
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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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Grimbal
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Re: Just Looking for an answer
« Reply #23 on: Nov 22nd, 2005, 5:27am » |
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Maybe that's why the problem statement asks for 2-18?
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Sjoerd Job Postmus
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Re: Just Looking for an answer
« Reply #24 on: Nov 22nd, 2005, 10:11am » |
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on Nov 22nd, 2005, 5:27am, Grimbal wrote:| Maybe that's why the problem statement asks for 2-18? |
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True, good point. But that doesn't make me not explore new possibilities.
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