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05 Jan 2024, 14:05
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:35) correct
42%
(02:42)
wrong
based on 53
sessions
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If P and Q are positive integers, then is (P + 2)(Q - 1) an even number?
(1) \(\frac{P}{3Q}\) is an even integer
(2) \(\frac{Q}{\sqrt{P}-3}\) is a positive odd integer
(1) \(\frac{P}{3Q}\) is an even integer
(2) \(\frac{Q}{\sqrt{P}-3}\) is a positive odd integer
05 Jan 2024, 20:56
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Bunuel
(P+2)(Q-1) is even when at least one of the two terms are even.
P+2 is even when P is even.
Q-1 is even when Q is odd.
Thus, we have to look for whether P is even and whether Q is odd.
(1) \(\frac{P}{3Q}\) is an even integer
\(\frac{P}{3Q}=2x\), where x is a positive integer.
\(P=6Qx\). This means P is even.
Sufficient
(2) \(\frac{Q}{\sqrt{P}-3}\) is a positive odd integer
\(\frac{Q}{\sqrt{P}-3}2x-1\)
\(Q=(\sqrt{P}-3)*(2x-1)\)
\(Q=(\sqrt{P}(2x-1)-3*(2x-1)\)
\(Q=(\sqrt{P}*odd-odd\)
Depends on P.
If P is even, then Q is odd, and if P is odd, then Q is even.
Examples: P=25, Q = 6 and P=16, Q=7
Insufficient
A
06 Jan 2024, 00:26
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Bunuel
P+2 is even if P is even and odd if P is odd
Q-1 is even if Q is odd and odd if Q is even
Valid combinations are
1. If P is even then all cases are true --- Eq(1)
2. P and Q are both odd --- Eq(2)
(1) \(\frac{P}{3Q}\) is an even integer
=> \(\frac{P}{3Q}\) = Even
=> P = 3Q * Even
=> P is even
As per Eq(1) this makes this validates that (P + 2)(Q - 1) is an even number always.
Option A is sufficient.
(2) \(\frac{Q}{\sqrt{P}-3}\) is a positive odd integer[/quote]
=> Q = Odd * ( \(\sqrt{P}-3\) )
Two Cases are possible
1. Q is even
Then \(\sqrt{P}-3\) has to be even
=> \(\sqrt{P}-3\) = even
=> \(\sqrt{P}\) = even + 3
=> \(\sqrt{P}\) = odd
=> P = \(odd^2\)
=> P = odd
Hence if Q is even, then P is odd --- This makes (P + 2)(Q - 1) an odd number hence we get a negative conclusion
1. Q is odd
Then \(\sqrt{P}-3\) has to be odd
=> \(\sqrt{P}-3\) = odd
=> \(\sqrt{P}\) = odd+ 3
=> \(\sqrt{P}\) = even
=> P = \(even^2\)
=> P = even
Hence if Q is odd, then P is even --- This makes (P + 2)(Q - 1) an even number hence we get a positive conclusion.
As a result, Option B is insufficient as it gives us valid and invalid conclusions.
IMO Option A
