Hardcore Maths Question
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((3628799 + 1) / 10) / (2 * 3 * 4 * 6) + (9 - 7 + 8 - 10) + 1 = 2519. ;)
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10! - 1 = 3628799. Its 1 less than a multiple of 1, 2, 3 ... 10.
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its the LCM of all the numbers (2520) - 1
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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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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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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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Jon Sagara wrote:
N(n) = 2519 + (n - 1)*(2520)
N(n) = (n * 2520) - 1 = 2520n - 1
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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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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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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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leppie wrote:
I wonder if its some kind of series...
It appears to be every 2520.
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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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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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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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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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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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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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2701439 last 4 digits by deduction, rest trial and error guess with 3/9 rule. :)
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It's not a mexican is it ?
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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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