I found a new theorem and I called it "My Remainder Theorem". My theorem finds similar thing with "Chinese Remainder Theorem". Maybe it has been founded before. If you heard this theorem before, please tell me.

Problem :

We have N equation.

x = equation_1.a ( mod equation_1.b )
x = equation_2.a ( mod equation_2.b )
.
.
.
x = equation_n.a ( mod equation_n.b )

(1<=i<=n) equation_i.b can be not a prime number. ( difference between my M.R.T. and C.R.T. )

We use a function which can do merge two equation and create a new equation. ( Let's call this function "merge" )

For example we have 3 equations. We'll solve this problem.

ans1 = merge( equation_1,equation_2 ); ans2 = merge( a,equation_3 );

Our answer is ans2.a.

Let's look at to "merge" function.

• We assume equation_1.b > equation_2.b
• lcm -> least common multiplier

merge( equation_1,equation_2 ){
	LCM = lcm(equation_1.b,equation_2.b)
	KKK = LCM/equation_1.b
	for i=0 to KKK-1
		if( ( i*equation_1.b + equation_1.a ) mod equation_2.b == equation_2.a )
			return answer = make_equation( LCM , i*equation_1.b+equation_1.a )	//	x = i*equation_1.b+equation_1.a(mod LCM)
	return "NO SOLUTION"
}

• If there is any solution for equation_1 and equation_2, also there is a solution for i<=KKK-1.
• We try all possible i values and we get a x value for equation_1. Then we check if x equals to equation_2.b+equation_2.a .
• If this condition is true, we find a x value. // x = i*equation_1.b+equation_1.a(mod LCM)

• If there is no x value, that means solution doesn't exist.

This algorithm's complexity is O( max( equation_i.b )*n )