If n is a positive integer, what is the value of n?

Question

(1) (n^2 + 6n + 16) is perfectly divisible by (n + 4).
(2) n > 2

IanStewart · Accepted Answer

Let's take the expression in the question, and try to make (n+4) a factor of as much of that expression as we can:

\begin{align}
n^2 + 6n + 16 &= n^2 + 4n + 2n + 16 \
&= n(n + 4) + 2n + 8 + 8 \
&= n(n+4) + 2(n + 4) + 8 \
&= (n+2)(n + 4) + 8
\end{align}

(n + 4) is clearly a divisor of the first term in this sum, so if the entire sum is divisible by (n+4), then (n+4) must also be a divisor of the second term in this sum, which is 8. And if n is a positive integer, and n+4 is a divisor of 8, then n must be 4. So Statement 1 is sufficient.

pushpitkc · Answer

Since n is a positive integer, n>0

2. (n^2 + 6n + 16) is perfectly divisible by (n + 4).
If n=2, 4+12+16(32) is not perfectly divisible by 
 If n=4, 16+24+16(56) is perfectly divisible by 8 .
If n=8, 64+48+16(128) is not perfectly divisible by 12
If n=12, 144+72+16(232) is not divisible by 16.

Hence, the value of n is 4.  Sufficient

2. n > 2
This is clearly not sufficient to give us an unique value of n (Option A)

mohshu · Answer

Bunuel   chetan2u

if n+4 is a factor of any quadartic.... which means when i substitute n = -4 in the quadartic,, the euqation shud give me 0

in this case,,it does not... plz explain

mohshu · Answer

hi ,,,i just want to clarify,,, in a quadratic equation,, 
if (n+4) is a factor then n= -4 shud give zero or not?

IanStewart · Answer

Yes, if x + 4 were a factor of some quadratic, then plugging x = -4 into that quadratic will give you a result of zero. So for example, x + 4 is a factor of this quadratic:

x^2 + 7x + 12 = (x + 4)(x + 3)

and if you plug x = -4 into this quadratic, the result will be zero.

But notice also: if x is a positive integer, then x^2 + 7x + 12 will  always  be divisible by x+4, because you can write the number x^2 + 7x + 12 as a product of two smaller integers, using the factorization above. In the question Bunuel posted above, n+4 is definitely  not  a factor of the quadratic n^2 + 6n + 16, because n^2 + 6n + 16 is not divisible by n+4 for every single value of n. It's only divisible by n+4 for certain values of n.

If that's confusing at all, it might help to think of much simpler situations: Say k is a positive integer:

* Algebraically speaking, k is a factor of 2k. So 2k is  always  divisible by k, no matter what k equals
* Algebraically speaking, k is not a factor of k + 2. But k could still sometimes be a divisor of k+2 for certain values of k. It turns out here that it is, only when k = 1 and k = 2.

pradeepmaria · Answer

Statement A:
(n^2 + 6n + 16) is perfectly divisible by (n + 4)
(n^2 + 6n + 16) = (n^2 + 8n + 16) - 2n
 = (n+4)^2 - 2n
Therefore 2n s a multiple of (n+4)

2n = n + 4; => n = 4
2n = 2(n + 4); => 0 = 4  Cant be true 
2n = 3(n + 4); => n = -12  n is positive 
The only way this is true is when n = 4
A is Suff

B clearly does not suffice

Ans: A

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If n is a positive integer, what is the value of n?

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If n is a positive integer, what is the value of n?

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