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 \(1\).
The OA is C
Source: Manhattan Prep
Remainders
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Let's take each statement one by one.swerve wrote: ↑Mon Feb 17, 2020 10:04 amEight 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 \(1\).
The OA is C
Source: Manhattan Prep
1. The remainder when the largest of the consecutive integers is divided by \(x\) is 0.
Case 1: Say the 8 consecutive integers are 1, 2, 3, 4, 5, 6, 7 and 8, and x = 2, then the remainders are 1, 0, 1, 0, 1, 0, 1, 0. Sum = 4.
Case 2: Say the 8 consecutive integers are 1, 2, 3, 4, 5, 6, 7 and 8, and x = 8, then the remainders are 1, 2, 3, 4, 5, 6, 7 and 0. Sum = 28.
No unique answer. Insufficient.
2. The remainder when the second largest of the consecutive integers is divided by \(x\) is \(1\).
=> The remainder when the largest of the consecutive integers is divided by \(x\) is 1 – 1 = 0.
This is the same information that we have in Statement 1. Insufficient.
(1) and (2) together
Say the 8 consecutive integers are n, (n + 1), (n + 2), ... (n + 7).
From (1), we have n + 7 = xq, where q = quetient
Thus, n = xq – 7;
And from (2), we have n + 6 = x(q – 1) + 1 => n = xq – x – 5
=> xq – 7 = xq – x – 5 => x = 2.
Thus, the sum of the remainders when each of the integers is divided by x (= 2) = 4. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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