ttly awesome math thingy?

Question

find all integers n so that the equation nx-12=3x has an integer solution for x

the second 1:
find the difference between the sum of the first 2003 even countin numbrs and the sum of the first 2003 odd countin numbers

yay!

Answer 1

PROBLEM 1:

nx - 12 = 3x

nx - 3x = 12
(n - 3)x = 12
x = 12 / (n - 3)

To have an integer solution n - 3 must be a multiple of 12.

n - 3 = 12k (where k is an integer)

n = 12k + 3

Plugging in k from the set {... -3, -2, -1, 0, 1, 2, 3 ... } you get the possible values of n:
{... -33, -21, -9, 3, 15, 27, 39, etc.}

PROBLEM 2:

Let's take the first two numbers. Even number is 2, odd number is 1. Difference is 1.

Now if we had the first two even (2, 4} and the first two odd {1, 3} the differences would be 2-1 and 4-3, this is 2.

Continue the pattern and it is obviously that you will have a difference of 1 for every number. Since you have 2003 numbers in each set, the answer is 2003.

Answer 2

nx-12=3x

x(n-3) =12
x = 12/(n-3)
n-3 = 1; 2; 3;4; 6;12
n ={4 ; 5 ; 6; 7; 9; 15 }

also n = {-1 ; 0 ; -3; -9 }