varun289 wrote:If positive integer y is equal to the sum of all the unique factors of the positive integer x, is |x-y| > 1?
x is not prime
x≠1
Statement 1: x is not prime
If x=1, then x has EXACTLY ONE unique factor: 1 itself.
Thus, y=1.
The result:
|x-y| = |1-1| = 0.
In this case, |x-y| < 1.
If x = non-prime other than 1, then x has AT LEAST 3 unique factors: x, 1, and k, where 1<k<x.
Thus, y = x+1+k.
The result:
|x-y| = |x - (x+1+k)| = |-(1+k)| = 1+k.
Since the least possible value of k is 2, 1+k ≥ 3.
In this case, |x-y| ≥ 3.
INSUFFICIENT.
Statement 2: x≠1
As shown above, if x = non-prime other than 1, then |x-y| ≥ 3.
If x = prime, then x has EXACTLY 2 unique factors: x and 1.
Thus, y = x+1.
The result:
|x-y| = |x - (x+1)| = 1.
In this case, |x-y| = 1.
INSUFFICIENT.
Statements combined:
As shown above, if x = non-prime other than 1, then |x-y| ≥ 3.
SUFFICIENT.
The correct answer is
C.
A few cases to illustrate that |x-y| ≥ 3 if x = non-prime other than 1:
If x=4, then y = 1+2+4 = 7.
In this case, |x-y| = |4-7| = 3.
If x=6, then y = 1+2+3+6 = 12.
In this case, |x-y| = |1-12| = 11.
If x=8, then y = 1+2+8 = 11.
In this case, |x-y| = |8-11| = 3.
In every case, |x-y| ≥ 3.
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