Eight consecutive integers are selected from the integers 1

Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?

Note that when a positive integer is divided by positive integer x, then possible remainders are 0, 1, 2, ..., (x-1). For example, when a positive integer is divided by 4, then the possible remainders are 0, 1, 2, or 3.

(1) The remainder when the largest of the consecutive integers is divided by x is 0. This implies that the largest integer is a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=1, then the sum of the remainders will be 0+0+0+0+0+0+0+0=0. Not sufficient.

(2) The remainder when the second largest of the consecutive integers is divided be x is 1. This implies that the second largest integer is 1 more than a multiple of x. Now, if eight consecutive integers are {1, 2, 3, 4, 5, 6, 7, 8} and x=2, then the sum of the remainders will be 1+0+1+0+1+0+1+0=4, but if x=6, then the sum of the remainders will be 1+2+3+4+5+0+1+2=18. Not sufficient.

(1)+(2) We have that:
7th largest integer when divided by x yields the remainder of 1;
8th largest integer when divided by x yields the remainder of 0.

Thus, we have that 1 is the largest remainder possible, which implies that x=2 (x-1=1 --> x=2). The sum of the remainders when any eight consecutive integers are divided by 2 is 1+0+1+0+1+0+1+0=4. Sufficient.

Answer: C.

Hope it's clear.