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23 Mar 2022, 02:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
86% (00:48) correct
14%
(00:33)
wrong
based on 85
sessions
History
Date
Time
Result
Not Attempted Yet
23 Mar 2022, 08:37
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Bunuel
Target question: Is integer p even?
Statement 1: 2p + 1 = an odd number
Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN
#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN
Let's first subtract 1 from both sides of the equation to get: 2p = (an odd number) - 1
1 is odd
Since an odd number minus an odd number is even, we can write: 2p = some even number
Since multiplying a number by 2 will turn any number into an even number, statement 1 is not sufficient.
For example, if p = 3, then 2p = 2(3) = 6, which is even. In this case, the answer to the target question is NO, p is not even
Conversely, if p = 4, then 2p = 2(4) = 8, which is even. In this case, the answer to the target question is YES, p is even
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: p/2 is even
Useful property: All even integers can be expressed as 2k, where k is an integer
So, let's rewrite statement 2 as follows: p/2 = 2k (for some integer k)
Now multiply both sides of the equation by 2 to get: p = 4k
Since 4 is EVEN, we know that 4k will be EVEN for all integer values of k.
In other words, we can be certain of that p is EVEN
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
25 Oct 2023, 05:49
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