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If n is a positive integer and (n + 1)(n + 3)
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Updated on: 11 Dec 2022, 23:29

1

Question Stats:

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

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

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 .

Re: If n is a positive integer and (n + 1)(n + 3)
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13 Oct 2021, 22:15

2

If n is a positive integer and (n + 1)(n + 3) is odd , that means n must be even= 2,4,6,8 .........etc.

When n=2 then (n + 1)(n + 3) = 15 and (n + 2)(n + 4) = 24 that value are divisible by : 3 , 6 & 8

When n=4 then (n + 1)(n + 3) = 35 and (n + 2)(n + 4) = 48 that value are divisible by : 3 , 6 & 8

When n=6 then (n + 1)(n + 3) = 63 and (n + 2)(n + 4) = 80 that value are divisible by : 8 & 16

When n=8 then (n + 1)(n + 3) = 99 and (n + 2)(n + 4) = 120 that value are divisible by only 6 & 8

Question mentioned " must be a multiple" not could be, so can see in above calculation 8 is a must be multiple of (n + 2)(n + 4)

So, Answer must be only " 8 "

When n=2 then (n + 1)(n + 3) = 15 and (n + 2)(n + 4) = 24 that value are divisible by : 3 , 6 & 8

When n=4 then (n + 1)(n + 3) = 35 and (n + 2)(n + 4) = 48 that value are divisible by : 3 , 6 & 8

When n=6 then (n + 1)(n + 3) = 63 and (n + 2)(n + 4) = 80 that value are divisible by : 8 & 16

When n=8 then (n + 1)(n + 3) = 99 and (n + 2)(n + 4) = 120 that value are divisible by only 6 & 8

Question mentioned " must be a multiple" not could be, so can see in above calculation 8 is a must be multiple of (n + 2)(n + 4)

So, Answer must be only " 8 "

Re: If n is a positive integer and (n + 1)(n + 3)
[#permalink]
11 Dec 2022, 22:53

Re: If n is a positive integer and (n + 1)(n + 3)
[#permalink]
11 Dec 2022, 23:30

Expert Reply

The explanations above are correct but the OA was wrong. It is D

Re: If n is a positive integer and (n + 1)(n + 3)
[#permalink]
14 Dec 2022, 18:38

Why are A and C also not correct?

If n is a positive integer and (n + 1)(n + 3)
[#permalink]
15 Dec 2022, 02:17

Expert Reply

The explanations above are perfect

specifically the one above https://gre.myprepclub.com/forum/if-n-i ... tml#p81202

As you can see sometimes the number is divisible by 3,6, and 8

Sometimes 6 and 8

Sometimes 8 and 16

It depends

The only thing we do know for SUR is that n is even and that is ALWAYS divisible by 8

I suggest you to look at out theory about divisibility and number theory https://gre.myprepclub.com/forum/gre-ma ... 29264.html

specifically the one above https://gre.myprepclub.com/forum/if-n-i ... tml#p81202

As you can see sometimes the number is divisible by 3,6, and 8

Sometimes 6 and 8

Sometimes 8 and 16

It depends

The only thing we do know for SUR is that n is even and that is ALWAYS divisible by 8

I suggest you to look at out theory about divisibility and number theory https://gre.myprepclub.com/forum/gre-ma ... 29264.html

gmatclubot

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