Hardcore Maths Question

[Andrew Bleakley]

It's not a mexican is it ?

[leppie]

Here's a wild guess: 123456789 ? [edit] ok I got the 9 bit :p [edit] -- modified at 18:37 Wednesday 26th July, 2006

[leppie]

2701439 last 4 digits by deduction, rest trial and error guess with 3/9 rule. :)

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[Judah Gabriel Himango]

Provided I've understood the question correctly, I think I've solved it. I kind of cheated though; I wrote a C# program that solves this:

        int start = 1;
        int divisor = 10;
        while (divisor >= 2)
        {
            if (start % divisor == divisor - 1)
            {
                divisor--;
            }
            else
            {
                start++;
                divisor = 10;
            }
        }

Soon as that loop exits, you've got your number, which happens to be 2519.

[leppie]

Judah Himango wrote:

I kind of cheated.

He did say "find a number" :) So there are more than one of these. I wonder if its some kind of series...

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[Judah Gabriel Himango]

yeah yeah :) Hey, I worked for it at least. :) *edit* oooh, misunderstood you there leppie. I though you were chiding me for solving it with code rather than brain. :) Yes, there are other numbers, it appears every 2520 iteration matches the criteria.

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[Raj Lal]

Judah Himango wrote:

I kind of cheated though

well thats ok , and of course there are more numbers but the fun is when you deduce how to do it It's the journey, not the destination

Omit Needless Words - Strunk, William, Jr.

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[Raj Lal]

yes thats true there are many , but if you think it might take a day to get the solution , without any computer help, but its worth

Omit Needless Words - Strunk, William, Jr.

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[Paul Conrad]

There was a question of if there were more number, yes, there are. Here is a modification of your code that shows others :)

#include using namespace std;

int main()
{
int start = 1;
int divisor = 10;
while ( start <1000000 ) // Or whatevery you want in signed 32-bit range
{
while (divisor >= 2)
{
if (start % divisor == divisor - 1)
{
divisor--;
}
else
{
start++;
divisor = 10;
}
}

    cout<

[Judah Gabriel Himango]

leppie wrote:

I wonder if its some kind of series...

It appears to be every 2520.

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[Paul Conrad]

leppie wrote:

So there are more than one of these. I wonder if its some kind of series...

Take a look at the modification of Judah's code that I posted. Your number is one of the numbers that come up :)

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I'd like to help but I am too lazy to Google it for you.

[Judah Gabriel Himango]

Quartz... wrote:

It's the journey, not the destination

Very true. I actually had fun writing a little piece of code to solve it, though, so it was the journey even still. :) I added some more code that added each match to a list box on a Windows Form. Then, after seeing how it froze up the UI, I did it on a background thread. Still, the UI thread would get flooded with matches, almost preventing it from painting, so I further chagned the code to only update during app idle. Voila, cool little WinForms program that solves it. :cool:

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[Paul Conrad]

Judah Himango wrote:

It appears to be every 2520.

It is. Modifying the your code that I modified and posted, shows this to be true :)

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I'd like to help but I am too lazy to Google it for you.

[Paul Conrad]

Judah Himango wrote:

Voila, cool little WinForms program that solves it. :cool:

That's cool. Mine is just a plain boring console app :->

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I'd like to help but I am too lazy to Google it for you.

[Jon Sagara]

N(1) = 2519 N(2) = 2519 + 2520 = 5039 N(3) = 2519 + 2520 + 2520 = 7559 N(4) = 2519 + 2520 + 2520 + 2520 = 10079 ... N(n) = 2519 + (n - 1)*(2520) No idea what the heck it means, though. Care to enlighten us mathematically-challenged folks?

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[leppie]

Jon Sagara wrote:

N(n) = 2519 + (n - 1)*(2520)

N(n) = (n * 2520) - 1 = 2520n - 1

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[leppie]

Some other interesting and useless observations:

1*2*3*4*5*6*7*8*9 is divisible by 2520 = 144
2520 is divisible by 2 * 3 * 5 * 7 = 12 (product of prime 1 - 9)
4 * 6 * 8 * 9 is divisible by 144 = 12 (product of 'non' prime 1 - 9)

:doh:

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[JenovaProject]

10! - 1 = 3628799. Its 1 less than a multiple of 1, 2, 3 ... 10.

[Ingo]

((3628799 + 1) / 10) / (2 * 3 * 4 * 6) + (9 - 7 + 8 - 10) + 1 = 2519. ;)

------------------------------ PROST Roleplaying Game War doesn't determine who's right. War determines who's left.

[Bassam Abdul Baki]

x = (i - 1) (mod i), 2 ≤ i ≤ 10. Thus, x = LCM(2, 3, 4, 5, 6, 7, 8, 9, 10) * k + LCM(2, 3, 4, 5, 6, 7, 8, 9, 10) - 1 LCM(2, 3, 4, 5, 6, 7, 8, 9, 10) = LCM(5, 7, 8, 9) = 23.32.5.7 = 2520. Thus, x = 2520k + 2519. min(x) = 2519. I had to redo it since I did it backwards.

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Last modified: Thursday, July 27, 2006 12:08:26 PM --