Solving 19200+23200=(21−2)200+(21+2)200
=2C0200(21)200×20+C2200(21)198×22....C200200(21)0×2200
Now we know that 212=441 is divisible by 49,
So rest of the terms will be divisible by 49
So lets consider for 2C200200(21)0×2200=2201
Now rewriting the term 2201=(23)67=(7+1)67
Now (1+7)67=C067×1+C1677+C26772.........
=1+67×7+49k
Now when dividing (1+7)67 by 49 we consider 1+67×7 as 49k is divisible by 49
Now 1+67×7=470 which when divided by 49 leaves remainder as 29.