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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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It's not a mexican is it ?
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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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2701439
last 4 digits by deduction, rest trial and error guess with 3/9 rule.
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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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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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There was a question of if there were more number, yes, there are. Here is a modification of your code that shows others
#include <iostream>
using namespace std;
int main()
{
int start = 1;
int divisor = 10;
while ( start <1000000 )
{
while (divisor >= 2)
{
if (start % divisor == divisor - 1)
{
divisor--;
}
else
{
start++;
divisor = 10;
}
}
cout<<start<<endl;
divisor=10;
start++;
}
return 0;
}
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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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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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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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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.
Tech, life, family, faith: Give me a visit.
I'm currently blogging about: Messianic Instrumentals (with audio)
The apostle Paul, modernly speaking: Epistles of Paul
Judah Himango
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Judah Himango wrote: Voila, cool little WinForms program that solves it.
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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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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Jon Sagara wrote: N(n) = 2519 + (n - 1)*(2520)
N(n) = (n * 2520) - 1
= 2520n - 1
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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)
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its the LCM of all the numbers (2520) - 1
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10! - 1 = 3628799. Its 1 less than a multiple of 1, 2, 3 ... 10.
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((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.
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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.
"People who want to share their religious views with you almost never want you to share yours with them." - Anonymous
Web - Blog - RSS - Math - LinkedIn
Last modified: Thursday, July 27, 2006 12:08:26 PM --
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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.
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How many Americans does it take to change a lightbulb ?
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0
these kinds of work are normally handled by mexicans non - americans in america
-- modified at 19:18 Wednesday 26th July, 2006
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