Problem 358

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guptavis
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Joined: Fri Nov 11, 2011 2:47 pm

Problem 358

Post by guptavis »

Problem 358 (View Problem)

Just observed and the eight leftmost zeroes have to come as the rightmost digits in one of the products. So the number will have 100000000 digits!!!
******56789 X _____ will end in {edited away by hk}
I believe one has to usethis to get anywhere close to an answer.

wow 3 people have already gotthe correct answer!!
Maybe i should try out the problems on page 1 before venturing out to these :lol: :lol:
Last edited by guptavis on Mon Nov 14, 2011 2:45 am, edited 1 time in total.
mdean
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Re: Problem 358

Post by mdean »

I've got a result that seems to not be working. The first and last digits appear to be correct, it appears to be cyclic, and my code to add up the digits appears to be working, but the answer is not being accepted. Is there someone I can PM on this?
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TripleM
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Re: Problem 358

Post by TripleM »

You can PM me, in a couple of weeks, after people who wish to solve the problem themselves can have the chance to do so :)
mdean
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Re: Problem 358

Post by mdean »

Well, so much for One in a Hundred...

Update: Well, I modified my program to not return once it found a value. Are you absolutely sure there is only 1 cyclic number meeting these crtiteria? I find 2.
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mdean
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Re: Problem 358

Post by mdean »

Actually, looks like one of the hare-brained tests I came up with in the past didn't quite pan out as I for some reason thought it would. With the test corrected, I find only one cyclic number.
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BostonBear
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Re: Problem 358

Post by BostonBear »

guptavis wrote:Just observed and the eight leftmost zeroes have to come as the rightmost digits in one of the products. So the number will have 100000000 digits!!!
******56789 X _____ will end in {edited away by hk}
I believe one has to usethis to get anywhere close to an answer.

wow 3 people have already gotthe correct answer!!
Maybe i should try out the problems on page 1 before venturing out to these :lol: :lol:

I stlll haven't solved this yet but part of the reason is my computer is chocking on the huge number. I think I have gleaned the way to solve this but haven't hit paydirt yet. The reason I'm posting this is because I feel this post is going to steer some people away from the possible correct answer. From my calculations, this number should have way MORE than 10**8 digits, in fact it should be in the realm of x * 10**8 digits with 2<x<9 (not trying to give away too much here. Just saying, I think if someones reasoning is flawed maybe HK should point that out. If my reasoning is flawed, please tell me.
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hk
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Re: Problem 358

Post by hk »

You're not requested to give the entire number, but only the sum of its digits.
So you need to find a way to find the sum of its digits.
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guptavis
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Re: Problem 358

Post by guptavis »

BostonBear wrote:
guptavis wrote:Just observed and the eight leftmost zeroes have to come as the rightmost digits in one of the products. So the number will have 100000000 digits!!!
******56789 X _____ will end in {edited away by hk}
I believe one has to usethis to get anywhere close to an answer.

wow 3 people have already gotthe correct answer!!
Maybe i should try out the problems on page 1 before venturing out to these :lol: :lol:

I stlll haven't solved this yet but part of the reason is my computer is chocking on the huge number. I think I have gleaned the way to solve this but haven't hit paydirt yet. The reason I'm posting this is because I feel this post is going to steer some people away from the possible correct answer. From my calculations, this number should have way MORE than 10**8 digits, in fact it should be in the realm of x * 10**8 digits with 2<x<9 (not trying to give away too much here. Just saying, I think if someones reasoning is flawed maybe HK should point that out. If my reasoning is flawed, please tell me.
I see it now, your reasoning is correct. now i think it could be as high as 7* 10**8.
but all that is besides the point, some other logic has to be used.
mario62
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Joined: Tue Feb 24, 2009 8:57 am

Re: Problem 358

Post by mario62 »

mdean wrote:I've got a result that seems to not be working. The first and last digits appear to be correct, it appears to be cyclic, and my code to add up the digits appears to be working, but the answer is not being accepted.
Have you found a fix?
I started today with the problem and found 2 solutions
one with 725****1 digits (incl. zeroes)
and the other with 725*****1 (1 digit more in the number of digits - very similar number of digits...)

and i tried to multiply my values with 1000 different values and checked the digit-sum and it is always the same.
and the number starts with 0(8)137 and ends in 56789.
i checked for overflows in arithmetic (i use my own bignumber class).

currently i am out of ideas.

ms.
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hk
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Re: Problem 358

Post by hk »

mario62 wrote:i checked for overflows in arithmetic (i use my own bignumber class).

currently i am out of ideas.

ms.
You don't need a bignumber class to solve this problem.
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Marcus_Andrews
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Re: Problem 358

Post by Marcus_Andrews »

mario62 wrote:
mdean wrote: i checked for overflows in arithmetic (i use my own bignumber class).
currently i am out of ideas.
ms.
358's solution is achievable through exploiting certain mathematical insights (and such is the case with most PE problems where brute force is infeasible). All I can tell you, without giving anything away, is that it would be a good idea to do some research on cyclic numbers first. Sounds like obvious advice, but it will really save you a lot of time and effort once you get a better idea of which mathematical properties can be leveraged, here.
EpicWisdom
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Re: Problem 358

Post by EpicWisdom »

Mathematical insights? So far, I've discovered enough to obtain the range of the digit count (through verifiable means) and the sum of the digits of a given cyclic, but I'm a bit stuck on the verification/generation of the cyclic in question (specifically, uniquely identifying information about the cyclic required to obtain the sum of the digits).

Sorry for being so vague, not sure what might cross the line of being a spoiler...
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