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Eight consecutive integers are selected from the integers 1

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Eight consecutive integers are selected from the integers 1

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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.

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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.

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Post
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.

OA C.
Say, the eight consecutive integers are n, (n + 1), (n + 2), (n + 3), (n + 4), (n + 5), (n + 6), (n + 7), where n + 7 ≤ 50

We have to get the sum when the eight consecutive integers are divided by x.

Let's take each statement one by one.

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

=> n + 7 = px, whether p is quotient

We don't have the unique value of p, thus, we can't get n and x and then the required sum. Insufficient.

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

=> n + 6 = (p -1)x + 1, whether (p -1) is quotient

Since in Statement (1), we have p as quotient when the largest integer is divided by x, we must have (p - 1) as quotient when the second largest integer is divided by x.

We don't have the unique value of p, thus, we can't get n and x and then the required sum. Insufficient.

(1) and (2) together

From (1), we have n + 7 = px and from (2), we have n + 6 = (p -1)x + 1.

From (1), n + 7 = px => n = px - 7.

Plugging-in the value of n in n + 6 = (p -1)x + 1, we get

px - 7 + 6 = (p -1)x + 1

px - 1 = px - x + 1

x = 2

With x = 2 as divisor, the remainder would be either 0 or 1. Any even integer divided by 2 will leave remainder = 0 and any odd integer divided by 2 will leave remainder = 1.

Among any 8 consecuitve integers, there will be 4 even and 4 odd integers.

Sum of remainders when 4 even integers are divided by 2 = 0 + 0 + 0 + 0 = 0;
Sum of remainders when 4 odd integers are divided by 2 = 1 + 1 + 1 + 1 = 4

Thus, the sum of remainder when 8 consecuitve integer are divided by 2 = 0 + 4 = 4. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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