- 01 Aug '07 18:33There are 2 Series (S1 & S2)

S1 begins at 10 and increases by 1

S2 begins at 6 and decreases by 1

Find the first whole number at which S1 is divisable by S2.

*No Trial and Error allowed*

10/6 = 1Mod4

11/5 = 2Mod1

12/4 = 3 Answer

Just find how the answer 3(12/4) is resolved based on the initial inputs 10 & 6 and how you know each series behaves. - 13 Aug '07 13:33The only problem with this is that once again you are you are using trial and error to find the first number less than seven divisable into 16. Can this be determined.

Also,Can your initial work be applied to a more complex initial statement;

102+n/21-2n - Once again finding the first n at which a whole number is produced. I'd presume 102+n is required to be multiplied by 2 - 14 Aug '07 17:04 / 1 edit

highest common factor of 6 and 10 is 2.*Originally posted by jonnifandango***There are 2 Series (S1 & S2)**

S1 begins at 10 and increases by 1

S2 begins at 6 and decreases by 1

Find the first whole number at which S1 is divisable by S2.

*No Trial and Error allowed*

10/6 = 1Mod4

11/5 = 2Mod1

12/4 = 3 Answer

Just find how the answer 3(12/4) is resolved based on the initial inputs 10 & 6 and how you know each series behaves.

6 minus 2 is 4

10 plus 2 is 12

4/12 is the answer - 23 Aug '07 16:03Uzless, i suppose that what jonnifandango wants is a formula to get the solution.

I imagine the solution would be N="whatever", where N is the number you add to S1 and S2, and whatever will only depend on S1 and S2.

I've started working it out with (10+n)/(6-n)=I , and i'm already stuck. I need to find out a proper definition for I..... - 23 Aug '07 17:23This is an example of a Diophantine equation, of the form:

(a-bz) + n(1+z) = 0

Where a, b are given and n, z are integers. I checked the Wolfram site and apparently there is a solution for linear Diophantine equations with 2 variables. I'll post the example when I get back from my meeting, but here's the link to the algorithm:

http://mathworld.wolfram.com/DiophantineEquation.html - 23 Aug '07 18:58

Oops, I guess this isn't really a linear equation. I'll try this little algorithm and see if it works anyway.*Originally posted by PBE6***This is an example of a Diophantine equation, of the form:**

(a-bz) + n(1+z) = 0

Where a, b are given and n, z are integers. I checked the Wolfram site and apparently there is a solution for linear Diophantine equations with 2 variables. I'll post the example when I get back from my meeting, but here's the link to the algorithm:

http://mathworld.wolfram.com/DiophantineEquation.html - 23 Aug '07 19:06 / 1 edit

Formulas are for the weak my yinyang!*Originally posted by yinyang78***Uzless, i suppose that what jonnifandango wants is a formula to get the solution.**

I imagine the solution would be N="whatever", where N is the number you add to S1 and S2, and whatever will only depend on S1 and S2.

I've started working it out with (10+n)/(6-n)=I , and i'm already stuck. I need to find out a proper definition for I.....

How's this:

a+[factor a/b] =x

b-[factor a/b] =y

My formula is predicated on there actually being 2 numbers that will end up divided into each other