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Difficulty:
45%
(medium)
Question Stats:
67% (01:56) correct
33%
(02:00)
wrong
based on 132
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If n is a positive integer, then is (n + 1)(n + 3) a multiple of 4?
(1) (n + 2)(n + 4) is odd.
(2) (n + 3)(n + 6) is even.
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(1) (n + 2)(n + 4) is odd.
(2) (n + 3)(n + 6) is even.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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shridhar786
Joined: 31 May 2018
Last visit: 08 Feb 2022
Posts: 322
Location: United States
Concentration: Finance, Marketing
Posts: 322
If n is a positive integer, then is (n + 1)(n + 3) a multiple of 4?
STATEMENT--(1)--(n + 2)(n + 4) is odd.
From here we know that
(n+2)--odd and (n+4)--odd
since--(n+2)--odd this implies n = odd--(same in n+4)
now,
(n+1) = odd+odd = even (2 is a factor of n+1 since it is even)
(n+3) = odd +odd = even
(n+1)(n+3) = has 4 as a factor since both (n+1) and (n+3) are even
hence it makes (n+1)(n+3) divisible by 4
so SUFFICIENT
STATEMENT--(2)--(n + 3)(n + 6) is even.
(n+3)-odd and (n+6)-even
n = even
if n=2--(n+3)-odd and (n+6)-even
but (n + 1)(n + 3) = odd*odd = 3*5 = odd
hence 4 is not a factor of (n + 1)(n + 3)---answer is NO
(n+3)-even and (n+6)-odd
n = odd
if n =3--(n+3)-even and (n+6)-odd
(n+1)(n+3) = even*even = 4*6 = even
since both (n+1) and (n+3) are even 4 is a factor of (n + 1)(n + 3)---answer is YES
so INSUFFICIENT
A is the correct answer
STATEMENT--(1)--(n + 2)(n + 4) is odd.
From here we know that
(n+2)--odd and (n+4)--odd
since--(n+2)--odd this implies n = odd--(same in n+4)
now,
(n+1) = odd+odd = even (2 is a factor of n+1 since it is even)
(n+3) = odd +odd = even
(n+1)(n+3) = has 4 as a factor since both (n+1) and (n+3) are even
hence it makes (n+1)(n+3) divisible by 4
so SUFFICIENT
STATEMENT--(2)--(n + 3)(n + 6) is even.
(n+3)-odd and (n+6)-even
n = even
if n=2--(n+3)-odd and (n+6)-even
but (n + 1)(n + 3) = odd*odd = 3*5 = odd
hence 4 is not a factor of (n + 1)(n + 3)---answer is NO
(n+3)-even and (n+6)-odd
n = odd
if n =3--(n+3)-even and (n+6)-odd
(n+1)(n+3) = even*even = 4*6 = even
since both (n+1) and (n+3) are even 4 is a factor of (n + 1)(n + 3)---answer is YES
so INSUFFICIENT
A is the correct answer
Bunuel
(1) \((n + 2)*(n + 4)\) is odd.
odd * odd = odd
\(n+2= odd ---> n= odd-2= odd\)
if n= odd then, (n+1) (n+3) either of the two will make the product a multiple of 4. you can try and plug in some numbers and check.
n=1,3,5,7..etc.
(2) (n + 3)*(n + 6) is even.
even* even= even
odd* even = even
if, \(n+3 = even\) ---->\( n = even-3 = odd\)
if, \(n+3 = odd\) ------> \(n = odd-3 = even\)
hence, B is insufficient.
Answer is A.
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