Problem 038

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spen
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Re: Problem 038

Post by spen » Sun Sep 05, 2010 8:23 pm

I still must be misunderstanding something. The solution is supposed to be the largest number, nine digits, permutation of 1..9, right? The largest number I get that meets all of the requirements given is given as one of the examples, and is not accepted as the right answer. I can generate 4 other candidates, each from a 3 digit number multiplied by (1, 2, 3). The two examples given are the highest and lowest numbers that I can generate, so I have 3 others.

What am I missing?

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Slaunger
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Re: Problem 038

Post by Slaunger » Sun Sep 05, 2010 9:04 pm

Hi spen,

It appears to me that you understand the problem correctly, but have overlooked some candidates in your analysis. I do not know how many concatenated pandigital products with three, three-digit numbers there are as those never really caught my interest in the problem.

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mury
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Re: Problem 038

Post by mury » Mon Nov 21, 2011 1:58 pm

People are talking about reverse order here (n,...,3,2,1), but as far as I understand the problem, we are looking for the closest to 999999999 number that can be formed as the concatenated product of an integer and one of the lists:
(1,2)
(1,2,3)
(1,2,3,4)
.............
(1,2,3,4,...,n)
Am I right?

TripleM
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Re: Problem 038

Post by TripleM » Mon Nov 21, 2011 8:17 pm

Yes. The problem used to be worded differently, thus the confusion in the earlier posts.

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Yamaneko
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Re: Problem 038

Post by Yamaneko » Fri Jan 25, 2013 10:03 pm

I don't want to give away anything, but I have to tell you that I loved working on this one! At first I got sidetracked taking something for granted that I shouldn't have, and I got one of the pandigitals in the example, which of course wasn't the answer. :) Realising my supposition was wrong made me come up with better, more flexible code and also made me aware that I was overlooking some cases that I really shouldn't have. This time I loved being wrong. Cheers! :D
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brunnock
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Re: Problem 038

Post by brunnock » Tue Feb 05, 2013 12:43 pm

I believe the title is misspelled. I think it should be "multiple".

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hk
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Re: Problem 038

Post by hk » Tue Feb 05, 2013 1:49 pm

Thanks, changed.
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dunne
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Re: Problem 038

Post by dunne » Fri Mar 22, 2013 1:39 pm

I am completely stuck on this. I can't for the life of me see why the second example is not the right answer -- and my program agrees with me! ;-) Does anyone feel like giving me a slight hint?
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thundre
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Re: Problem 038

Post by thundre » Fri Mar 22, 2013 2:26 pm

dunne wrote:I am completely stuck on this. I can't for the life of me see why the second example is not the right answer -- and my program agrees with me! ;-) Does anyone feel like giving me a slight hint?
The number of concatenated products can be anywhere from 2 to 9. Don't get in a rut thinking that they must all be the same number of digits.
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dunne
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Re: Problem 038

Post by dunne » Fri Mar 22, 2013 3:07 pm

thundre wrote:
dunne wrote:I am completely stuck on this. I can't for the life of me see why the second example is not the right answer -- and my program agrees with me! ;-) Does anyone feel like giving me a slight hint?
The number of concatenated products can be anywhere from 2 to 9. Don't get in a rut thinking that they must all be the same number of digits.
Thanks, thundre. I am sure I have avoided this problem. Here's what I am doing (hopefully not giving away too much -- I doubt it, since my method is clearly flawed!) :

I am taking from 1 to 4 (although I think 3 is the actual limit) of the most signficant digits in turn from each pandigital, and generating and concatenating the products one by one, using 1,2,3...9. I don't bother about the number of products or the number of digits in each product. I have a check that stops this generation/concatenation as soon as the pandigital is reached or exceeded. I am doing this for *all* pandigitals, starting with the biggest. I get valid results, but the highest is always the second example from the problem text.
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thundre
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Re: Problem 038

Post by thundre » Fri Mar 22, 2013 5:49 pm

dunne wrote: I am taking from 1 to 4 (although I think 3 is the actual limit) of the most signficant digits in turn from each pandigital, and generating and concatenating the products one by one, using 1,2,3...9. I don't bother about the number of products or the number of digits in each product. I have a check that stops this generation/concatenation as soon as the pandigital is reached or exceeded. I am doing this for *all* pandigitals, starting with the biggest. I get valid results, but the highest is always the second example from the problem text.
So you're checking all 9! pandigitals, or at least the ~8! which are greater than 918273645.

If instead of a pandigital you give it 666613332, does it acknowledge that it's 6666 * (1,2)?
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dunne
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Re: Problem 038

Post by dunne » Fri Mar 22, 2013 6:50 pm

thundre wrote: So you're checking all 9! pandigitals, or at least the ~8! which are greater than 918273645.

If instead of a pandigital you give it 666613332, does it acknowledge that it's 6666 * (1,2)?
I am *generating* all 9!, but since I sort before checking and start checking at the end of the array, I stop at the first one found, which by definition is the biggest.

I gave it that number, it *didn't* acknowledge it, and almost at once I knew exactly why! Thanks a lot thundre, just a simple coding error but without talking it through with someone I doubt I would have spotted it any time soon.
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ubershmekel
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Problem 038

Post by ubershmekel » Mon Apr 30, 2018 7:31 am

Take the number 192 and multiply it by each of 1, 2, and 3:
[..]
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
When I read through this problem, I didn't realize what "1 to 9 pandigital" meant until I submitted the wrong answer a few times. My impression was that "1 to 9 pandigital" meant a number with 9 digits that was created through that multiplication form. I would recommend revising the text to define pandigital. For example:
By concatenating each product we get the 1 to 9 pandigital (a number that contains all the digits 1-9), 192384576...

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hk
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Re: Problem 038

Post by hk » Mon Apr 30, 2018 10:40 am

Please don't start a new topic for a problem if there exists one already.
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fistuk
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Re: Problem 038

Post by fistuk » Sun Nov 11, 2018 11:11 am

sorry couldn't find the other post so I'm asking here:

I probably don't get something regarding pandigitals since I had the same problem at 032.

I get a better answer that only the 2nd best is being accepted.
So maybe can someone explain me what's bad with 2469 *4 and 2469 *5 ?

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hk
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Re: Problem 038

Post by hk » Sun Nov 11, 2018 11:26 am

If you want to go up to 5 you have to concatenate 2469*1 and 2469*2 and 2469*3 and 2469*4 and 2469*5.
This gives 246949387407987612345
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