Data Sufficiency

Manhattan Prep

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?

1) The remainder when the largest of the consecutive integers is divided by x is 0.
2) The remainder when the second largest of the consecutive integers is divided by x is 1.

OA C.

AAPL wrote:Manhattan Prep

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?

1) The remainder when the largest of the consecutive integers is divided by x is 0.
2) The remainder when the second largest of the consecutive integers is divided by x is 1.

When consecutive positive integers are divided by positive integer x, the resulting remainders also are consecutive:
If x=4:
16/4 = 4 R0.
17/4 = 4 R1.
18/4 = 4 R2.
19/4 = 4 R3.
20/4 = 5 R0.
From here, the cycle of remainders will repeat: 0,1,2,3,0,1,2,3...

The smallest remainder is 0.
The greatest remainder is x-1 = 3. (This is a rule of remainders, regardless of the value of x: when positive integer y is divided by positive integer x, the greatest possible remainder is x-1).
Note also that the greatest possible remainder occurs right before the cycle begins again with R=0.

Statement 1: The remainder when the largest of the consecutive integers is divided by x is 0.
Tells us only that the largest integer is a multiple of x.
It's possible that the integers are 9,10,11,12,13,14,15,16 and that x=2, yielding remainders of 1,0,1,0,1,0,1,0.
Sum = 1+0+1+0+1+0+1+0 = 4.
It's possible that the integers are 9,10,11,12,13,14,15,16 and that x=1, yielding remainders of 0,0,0,0,0,0,0,0.
Sum = 0+0+0+0+0+0+0+0 = 0.
INSUFFICIENT.

Statement 2: The remainder when the second largest of the consecutive integers is divided by x is 1.
Tells us only that the second largest integer is 1 more than a multiple of x.
It's possible that the integers are 9,10,11,12,13,14,15,16 and that x=2, yielding remainders of 1,0,1,0,1,0,1,0.
Sum = 1+0+1+0+1+0+1+0 = 4.
It's possible that the integers are 10,11,12,13,14,15,16,17 and that x=5, yielding remainders of 0,1,2,3,4,0,1,2.
Sum of the remainders = 0+1+2+3+4+0+1+2 = 13.
INSUFFICIENT.

Statements 1 and 2 combined:
When the 7th integer is divided by x, R=1.
When the 8th integer is divided by x, R=0.
Since the greatest possible remainder occurs right before the cycle begins again with R=0, the greatest possible remainder here is 1.
Since the smallest possible remainder is 0, only one possible cycle of remainders is possible: 1,0,1,0,1,0,1,0.
Sum = 1+0+1+0+1+0+1+0 = 4.
SUFFICIENT.

The correct answer is C.

When the statements are combined:
Since the greatest possible remainder is x-1 = 1, we know that x=2 and the greatest of the 8 consecutive integers is even.
To illustrate:
When {9,10,11,12,13,14,15,16} are divided by 2, the remainders are 1,0,1,0,1,0,1,0.
When {23,24,25,26,27,28,29,30} are divided by 2, the remainders are 1,0,1,0,1,0,1,0.