Impossible or Easy?

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.

We have to solve: (n+9)/(-n+7)=integer.

But (n+9)/(-n+7)=-1+16/(-n+7).

-1 is integer, so (n+9)/(-n+7)=integer iff 16/(-n+7) is integer.

the biggest divisor of 16 smaller than 7 is 4. so -n+7=4, n=3, and the nubers are 12 and 4.

The 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

(n+102)/(-2n+21) = integer

(-0.5)(-2n+21)+112.5/(-2n+21) = integer ... Multiply by 2

2n-21+225/(-2n+21) = even integer

225/(-2n+21) = odd integer

225 = 15^2=3^2*5^2, so -2n+21=15, n=3

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.

no one debunked my solution?

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.....🙄

I should have really started with (S1 + N)/(S2 -N)=I

What can i replace "I" for?

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

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

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.....🙄