Hardcore Maths Question

Raj Lal · xacc.ide-0.2.0.50 - now with partial MSBuild support!

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

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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 · xacc.ide-0.2.0.50 - now with partial MSBuild support!

leppie wrote:

I wonder if its some kind of series...

It appears to be every 2520.

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Paul Conrad · xacc.ide-0.2.0.50 - now with partial MSBuild support!

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 :)

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 :)

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 :->

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. ;)

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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 --

Kacee Giger

I know I'm a little late (and a valid solution has already been given), but no real explanation has been made. You first need to find a number that is a multiple of all these multiples (10 * 9, 9 * 8, etc), then one less than that will give the proper remainders. So, to find the least common multiple, first break these into primes: 10 * 9 = 2 * 3 * 3 * 5, 9 * 8 = 2 * 2 * 2 * 3 * 3, 8 * 7 = 2 * 2 * 2 * 7, etc. Take out what is unique for each to get 2 * 2 * 2 * 3 * 3 * 5 * 7 = 2520. So, one answer (though you already know) to the original problem is 2519.