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05 Jun 2020, 04:10
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:26) correct
41%
(02:24)
wrong
based on 162
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If n is a positive integer, is n divisible by 2?
(1) 7n - 8 is divisible by 20
(2) 3n^2 + 2n + 5 is a prime number
Are You Up For the Challenge: 700 Level Questions
(1) 7n - 8 is divisible by 20
(2) 3n^2 + 2n + 5 is a prime number
Are You Up For the Challenge: 700 Level Questions
ArunSharma12
Joined: 25 Oct 2015
Last visit: 20 Jul 2022
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Updated on: 05 Jun 2020, 06:17
Originally posted by ArunSharma12 on 05 Jun 2020, 04:36.
Last edited by ArunSharma12 on 05 Jun 2020, 06:17, edited 1 time in total.
Last edited by ArunSharma12 on 05 Jun 2020, 06:17, edited 1 time in total.
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Bunuel
Given: n > 0
statement 1: 7n - 8 = 20k
7n = 20k + 8
7n = 7(2k + 1) + 6k + 1
n = (2k+1) + (6k+1/7)
6k+1/ 7 can only be odd
n will be even
sufficient
statement 2:
3n^2 + 2n + 5 = prime
let, 3n^2 + 2n + 5 = 2
3n^2 + 2n + 3 = 0; but n will not be a positive integer
3n^2 + 2n + 5 = odd prime
n has to be even and hence divisible by 2
sufficient
Ans:
firas92
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05 Jun 2020, 06:10
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Bunuel
(1) \(7n - 8\) is divisible by \(20\)
\(7n-8=20a\) or \(7n=20a+8=4(5a+2)\)\(\)
\(n=\frac{4(5a+2)}{7}\)
Since \(n\) is an integer and \(4\)\(\) is not multiple of \(7\), \((5a+2)\) must be a multiple of \(7\) and so \(n\) must be a multiple of \(4\)
So \(n\) is even and divisible by \(2\)
1 is sufficient
(2) \(3n^2 + 2n + 5\) is a prime number
Minimum value of \((3n^2 + 2n + 5)\) is \(10\) so \((3n^2 + 2n + 5)\) must be a prime number greater than \(10\) and so \((3n^2 + 2n + 5)\) must be odd
If \(n\) is odd, \((3n^2 + 2n + 5) = odd+even+odd = even\) ---------- Not possible since we know that \((3n^2 + 2n + 5)\) is odd
So \(n\) is not odd which means \(n\) is even and divisible by \(2\)
2 is sufficient
Answer is (D)
