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Hardcore Maths Question

Algorithms

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R Raj Lal

its the LCM of all the numbers (2520) - 1

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J 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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J 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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L 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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L leppie

Jon Sagara wrote:

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

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

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P 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.
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P 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<
```
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P 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 :)

I'd like to help but I am too lazy to Google it for you.
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J Judah Gabriel Himango

leppie wrote:

I wonder if its some kind of series...

It appears to be every 2520.

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R 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

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P 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.
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R 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

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J 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.
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J 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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L 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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L leppie

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

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A Andrew Bleakley

It's not a mexican is it ?
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R Raj Lal

Ok lets me be the first to ask a maths question Find a number which 1. divided by 10 gives a remainder 9 2. divided by 9 gives remainder 8 --- --- so on till divided by 2 gives a remainder 1 Any one ?

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L leppie

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