GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If n is a positive integer and (n + 1)(n + 3)
[#permalink]
Updated on: 11 Dec 2022, 23:29
Updated on: 11 Dec 2022, 23:29
1
00:00
Question Stats:
33% (02:21) correct
66% (01:07) wrong based on 30 sessions
Hide Show timer Statistics
If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?
(A) 3
(B) 5
(C) 6
(D) 8
(E) 16
(A) 3
(B) 5
(C) 6
(D) 8
(E) 16
Re: If n is a positive integer and (n + 1)(n + 3)
[#permalink]
01 Jun 2018, 15:31
01 Jun 2018, 15:31
1
Expert Reply
shahul wrote:
If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?
(A) 3 (B) 5 (C) 6 (D) 8 (E) 16
(A) 3 (B) 5 (C) 6 (D) 8 (E) 16
For \((n + 1)(n + 3)\) to be odd \((n+1)\) is odd and \((n+3)\) is also odd.
Since only \(odd \times odd=odd\). So \(n\) must be even since only \(even + odd = odd\).
\((n + 2)(n + 4)= 2 \times (\frac{n}{2}+1) \times 2 \times (\frac{n}{2}+2)\).
Either \((\frac{n}{2}+1)\) is even or \((\frac{n}{2}+2)\) is even.
So this means there must be a factor of 8 (\(2 \times 2 \times 2\)).
Re: If n is a positive integer and (n + 1)(n + 3)
[#permalink]
12 Jun 2019, 07:10
12 Jun 2019, 07:10
sandy wrote:
shahul wrote:
If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?
(A) 3 (B) 5 (C) 6 (D) 8 (E) 16
(A) 3 (B) 5 (C) 6 (D) 8 (E) 16
For \((n + 1)(n + 3)\) to be odd \((n+1)\) is odd and \((n+3)\) is also odd.
Since only \(odd \times odd=odd\). So \(n\) must be even since only \(even + odd = odd\).
\((n + 2)(n + 4)= 2 \times (\frac{n}{2}+1) \times 2 \times (\frac{n}{2}+2)\).
Either \((\frac{n}{2}+1)\) is even or \((\frac{n}{2}+2)\) is even.
So this means there must be a factor of 8 (\(2 \times 2 \times 2\)).
---------------------------------------
From the question stem, n is positive
and further logic shows that n is even.
Therefore, (n+2)(n+4) must be even.
the above entails as n^2+6n+8
as n is positive and even, we can, for sure, assume that n must be equal to 2 [ n may be other values like 4,6,8,10, etc]
if we take n=2 and plug in the above equation we 24 which is a multiple of 3,6,and 8.
Kindly correct my understanding of the above. Regards .
