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kaladin123
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30 May 2021, 01:18
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It took me some time to figure this one out. I think this is a high-quality question.
Am sharing my approach:
\(x=3y+z\)
\(y>z\) (divisor > remainder)
\(z=qy+2\)
\(y>2\)
\(x=3y+qy+2=(3+q)y+2\)
Now, \(y=\){3,4,5,...} (x, y are positive integers)
\(x_{min}=(3+q)(3)+2=(3+0)*3+2=9+2=11\)
I & II are eliminated - A, B, D, & E are eliminated.
Hence C.
Am sharing my approach:
\(x=3y+z\)
\(y>z\) (divisor > remainder)
\(z=qy+2\)
\(y>2\)
\(x=3y+qy+2=(3+q)y+2\)
Now, \(y=\){3,4,5,...} (x, y are positive integers)
\(x_{min}=(3+q)(3)+2=(3+0)*3+2=9+2=11\)
I & II are eliminated - A, B, D, & E are eliminated.
Hence C.
Flypaper3533
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GMAT 1: 740 Q48 V44
GMAT 2: 770 Q49 V47 (Online)
GPA: 3.98
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Schools: Kelley '25 (A$$$$) Johnson (Cornell) (WL) Emory 1YR '24 (A$$$) McCombs '25 (A$) Marshall '25 (A$$$$)
GMAT 2: 770 Q49 V47 (Online)
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24 Oct 2022, 00:51
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While the official solution is great for understanding the concept, I feel trial and error is easier in this case.
x, y are positive integers. Since quotient is >0, we know 0 < y < x.
I. x=5. There is no positive integer less than 5 that divides 5 and leaves a quotient of 3. Cannot be true.
II. x=8. There is no positive integer less than 8 that divides 8 with quotient =3. With y=2, quotient is 4. With y=3, quotient is 2. Cannot be true.
We can now rule out every option except 'C'.
x, y are positive integers. Since quotient is >0, we know 0 < y < x.
I. x=5. There is no positive integer less than 5 that divides 5 and leaves a quotient of 3. Cannot be true.
II. x=8. There is no positive integer less than 8 that divides 8 with quotient =3. With y=2, quotient is 4. With y=3, quotient is 2. Cannot be true.
We can now rule out every option except 'C'.
