Bunuel wrote:
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.
Did you just happen to know that the only way two consecutive integers can have this happen is if they're both divided by 2? I'm curious where you know that from, and if there's more rules like that which I could learn.