Find the divisor that is divisible by all these three numbers that has the same remainder.: 3000, 2913, 3203?

Answer 1

Let's see if we get this right.

The number is N
when N is divided by 3000, it has a remainder of k.
when N is divided by 2913, it has a remainder of k
when N is divided by 3203, it has a remainder of k

Therefore (N-k) is divisible by all three.

3000 = 2*2*2*3*5*5*5
2913 = 3*971
3203 = 3203 (it is prime)

(N-k)= 3000*971*3203 = 9,330,339,000
k could be 1
N = 9,330,339,001

Answer 2

Rephrase: 3000=amod(d)

2913=amod(d)

3203=amod(d)

This means that the differences = 0 mod(d)

Therefore, d divides 87,290, and 203

and forces d=29.