GMATT73 wrote:

Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

Surely duttsit's approach is the shortest one. In case, one doesn't know those mod rules, here is the way:

Let A be the divisor, X and Y be the two numbers. ( A is an integer)

We have: X= Ax + 431

Y= Ay+ 379

X+Y= Az + 211= A(x+y) + 810 ( x, y and z and integers)

---> A ( z-x-y)= 810-211= 599

Since 599 is a prime number as duttsit pointed out, we have two cases:

A= 599, z-y-x=1 ( the vice versa is impossible coz if A=1, there won't be any remainder since every number is divisible by 1)

A= -599, z-x-y=-1

Uhm, I think the negative value is valid...