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01 May 2019, 07:52
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:00) correct
67%
(02:26)
wrong
based on 81
sessions
History
Date
Time
Result
Not Attempted Yet
Which of the following expressions is never a prime number when p is a prime number?
(A) p^2 +16
(B) p^2 +24
(C) p^2 +26
(D) p^2 +46
(E) p^2 +96
(A) p^2 +16
(B) p^2 +24
(C) p^2 +26
(D) p^2 +46
(E) p^2 +96
03 May 2019, 05:59
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Noshad
Which of the options would never be a prime
That means if we can prove one of options as prime we can eliminate it.
That means we won't be able to take anything common between the two terms
So Let's start with eliminating the options
A) 24+9=33,
24+49=73 Prime Eliminate
B) 16+9=25
16+49=65
16+121= 137 Prime Eliminate
C) 25+26=51
49+26= 75
121+26=147
We ll come later
D) 25+46=71 Prime Eliminate
E) 25+96=121
49+96=143
121+96=217 Prime Bingo Eliminate
C is our answer
P.S Note All these calculations should run in our mind and we should never try 4 and never try 49 after trying 9 as the summation would always be a multiple of 5.
Being smart will save us some time in Gmat.
Good question Noshad
03 May 2019, 06:28
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If p is a prime greater than 3, then p cannot be divisible by 3 (since it's prime). So the remainder will either be 1 or 2 when we divide p by 3. But in either case, when we divide p^2 by 3, the remainder will be 1 -- you can just square the remainders themselves, to get 1 or 4, then take the remainder by 3, to get 1 or 1. So p^2 = 3q + 1 for some integer q, and p^2 + 26 = 3q + 1 + 26 = 3q + 27, which is divisible by 3 and is not prime.
So if the prime p > 3, then p^2 + 26 is never prime. The last thing we need to check is what happens if p is not greater than 3, i.e. if p = 2 or 3. In each case, p^2 + 26 is not prime, so C is the correct answer.
It's not hard to get down to C or E just using process of elimination (plugging in p=5 and 7 gets you there). But to demonstrate that p^2 + 96 can be prime, you actually need to plug in a value for p of at least 19, and when you do that, you have to confirm that 19^2 + 96 = 457 is a prime number, which is a very time-consuming thing to do. The solution above which uses p=11 to eliminate answer E is incorrect, since 11^2 + 96 = 217, and 217 is divisible by 7 and 31, and is not a prime number.
So if the prime p > 3, then p^2 + 26 is never prime. The last thing we need to check is what happens if p is not greater than 3, i.e. if p = 2 or 3. In each case, p^2 + 26 is not prime, so C is the correct answer.
It's not hard to get down to C or E just using process of elimination (plugging in p=5 and 7 gets you there). But to demonstrate that p^2 + 96 can be prime, you actually need to plug in a value for p of at least 19, and when you do that, you have to confirm that 19^2 + 96 = 457 is a prime number, which is a very time-consuming thing to do. The solution above which uses p=11 to eliminate answer E is incorrect, since 11^2 + 96 = 217, and 217 is divisible by 7 and 31, and is not a prime number.
