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m, n and k are positive integers. If the

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m, n and k are positive integers. If the  [#permalink]

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New post 16 May 2018, 11:03
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Question Stats:

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m, n and k are positive integers. If the product mn is odd, is k odd?

(1) mn + n + k is odd
(2) n^2 – kn – 6k^2 is even

*Kudos for all correct solutions

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m, n and k are positive integers. If the  [#permalink]

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New post 16 May 2018, 11:50
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GMATPrepNow wrote:
m, n and k are positive integers. If the product mn is odd, is k odd?

(1) mn + n + k is odd
(2) n^2 – kn – 6k^2 is even

*Kudos for all correct solutions


\(m*n=Odd\) implies that \(m=Odd\) & \(n=Odd\)

Statement 1: \(mn+n+k=odd\)

\(=> Odd+Odd+k=Odd => k=Odd\). Sufficient

Statement 2: \(n^2-kn-6k^2=Even\)

\(=> Odd-k*Odd-Even=Even\)

Therefore \(k*Odd=Odd => k=Odd\). Sufficient

Option D
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Re: m, n and k are positive integers. If the  [#permalink]

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New post 16 May 2018, 11:53
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GMATPrepNow wrote:
m, n and k are positive integers. If the product mn is odd, is k odd?

(1) mn + n + k is odd
(2) n^2 – kn – 6k^2 is even

*Kudos for all correct solutions


m*n is odd it means both m and n has to be odd.

stmt-1: mn + n + k = odd
as we know m and n is odd then we have
odd + odd + k = odd
even + k = odd
now k has to be odd here. so sufficient.

stmt-2:
n^2 -kn - 6k^2 = even
n^2 -k(n+6k) = even
n^2 - even = k(n+6k)
as we know n is odd, n^2 is also odd, so
odd - even = k(n+6k)
odd = k(n+6k) (because odd - even = odd)

for k*(n+6k) to be odd both k and n+6k has to be odd. so we know k is odd.
Sufficient.
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Re: m, n and k are positive integers. If the  [#permalink]

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New post 18 May 2018, 10:10
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1
GMATPrepNow wrote:
m, n and k are positive integers. If the product mn is odd, is k odd?

(1) mn + n + k is odd
(2) n² – kn – 6k² is even

*Kudos for all correct solutions


Target question: Is k odd?

Given: The product mn is odd
If the product mn is odd, when we know that m is ODD and n is ODD

Statement 1: mn + n + k is odd
In other words: (ODD)(ODD) + ODD + k is odd
Simplify: ODD + ODD + k is odd
Simplify more: EVEN + k is odd
This means k must be ODD
So, the answer to the target question is YES, k IS odd
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: n² – kn – 6k² is even
Let's factor the expression to get: (n - 3k)(n - 2k) is even
In other words: (ODD - 3k)(ODD - 2k) is even
Notice that 2k must be EVEN, so we can write: (ODD - 3k)(ODD - EVEN) is even
ODD - EVEN = ODD, so we can now say: (ODD - 3k)(ODD) is even
This tells us that (ODD - 3k) must be EVEN
In order for (ODD - 3k) to be EVEN, it must be the case that 3k is ODD
If 3k is ODD, then k must be ODD
So, the answer to the target question is YES, k IS odd
Since we can answer the target question with certainty, statement 2 is SUFFICIENT


Answer: D

Cheers,
Brent
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Re: m, n and k are positive integers. If the   [#permalink] 18 May 2018, 10:10
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