If in a long division sum,the dividend is 380606 and the successive remainders from the first to the last are 434, 125 and 413, then the divisor is?

Let d = divisor and q = quotient

First remove the last remainder:
d x q + 413 = 380606
d x q = 380193

So d cannot be even (choice d) or a multiple of 5 (choice c).

If we try choice a or b,
a quick division (by calculator) shows that
380193 = 451 x 843, the other two choices.

(In the absence of choices, we could find the prime factors of 380193,
which are 3 x 11 x 41 x 281 and get various candidate divisors from that.)

So one of them is the divisor and the other the quotient.

So we can just check the first remainder:
843 x 4 = 3372
3806 - 3372 = 434 ... a match!

And just to be sure:
451 x 8 = 3608
3806 - 3608 = 198

So the divisor is 843.