Danny's divisors

Using polynomial division one can determine that

$\frac{n^{2} + 40}{n + 4} = n - 4 + \frac{56}{n + 4}$

This is the factor by which $n^{2} + 40$ is greater than $n + 4$

For n to be an integer $n + 4$ must be a factor of 56

The factors of 56 are $1, 2, 4, 7, 8, 14, 28$ and $56$

Equating $n +4$ to each of the multiple of 56 gives us the list of possible integer solutions of n

$n = 52, 24, 10, 4, 3, 0, -2$ or $-3$

Rejecting solutions of n that are not positive integers we determine

$n = 52, 24, 10, 4$ or $3$

Summing the possible solutions of n obtains

$52 + 24 + 10 + 4 + 3 = 93$

By dividing $n^{2} + 40$ by $n + 4$ ,

We have $n^{2} + 40 = (n + 4)(n - 4) + 56$ So, for $n^{2} + 40$ to be divisible by $n + 4$ , 56. must be divisible by the latter.

56 has the following factors : 1,2,4,7,8,14,28,56.

So, $n + 4$ can be 7,8,14,28,56.

$n$ = 3,4,10 24,52.

So our answer is $3 + 4 + 10 + 24 + 52 = \boxed{93}$

We want $\frac {n^2+40}{n+4}$ to be an integer, so by long division we get $n-4+\frac {56}{n+4}$, which will be an integer whenever $frac {56}{n+4}$ is. We check that the factors of $56$ are $1, 2, 4, 7, 8, 14, 28, 56$ and we set $n+4$ equal to each of these factors. We add up all the positive values for $n$ to get $93$

Because $n^2 + 40$ is a multiple of $n+4$ , we can write this as the equation

$n^2 + 40 = k(n+4)$ , for some integer constant $k$ .

Dividing both sides of the equation by $n+4$ , we get

$\frac{n^2+40}{n+4} = k$ .

By dividing out the left-hand side of the equation, we get

$n-4 + \frac{56}{n+4} = k$ .

Remember that k has to be an integer. Since we can assume from the problem statement that $n$ is an integer, we only need to ensure that $\frac{56}{n+4}$ is an integer to ensure that $k$ is an integer as well. To do this, we look at all of the factors of 56, namely 1, 2, 4, 7, 8, 14, 28, and 56, and set these all equal to $n+4$ . By doing this, we get the values of $n$ for which $\frac{56}{n+4}$ reduces to an integer. Adding up all the values of $n$ , remembering that $n$ has to be positive, gives us an answer of 93.

n+4│n2 + 40, we know that n+4│n2 -16, so n+4│n2 + 40 –(n2 -16) , and we get n+4│56, so 56 is divisible by n+4, then n+4 is factors of 56,

case 1 56 = 1 X 56

If n+4 = 1 → n = -3 , If n + 4 = 56 → n = 52

Case 2 56 = 2 X 28

If n + 4 = 2 → n = -2, If n + 4 = 28 → n = 24

Case 3 56 = 4 X 14

If n + 4 = 4 → n = 0, If n + 4 = 14 → n = 10

Case 4 56 = 7 X 8

If n + 4 = 7 → n = 3, If n + 4 = 8 → n = 4

the sum of all positive integers n is 3 + 4 + 10 + 24 + 52 = 93

(n^2 + 40)/(n+4) = [(n+4)^2 + 24 – 8n]/(n+4) = [(n+4)^2 – 8(n-3)]/(n+4)

= (n+4) – 8(n-3)/(n+4)

= (n+4) – 8[(n+4)-7]/(n+4) = (n+4) – 8(1 – 7/(n+4)) = (n+4) – 8 + 56/(n+4) = (n -4) + 56/(n+4)

n<=52

2/(n+4) , 4/(n+4), 8/(n+4), 14/(n+4), 28/(n+4), 56/(n+4)

2^3*7

Multiple of 56

1,2,4,7,8,14,28,56

4, 7, 8, 14, 28, 56

4/(n+4) = n=0. (not included)

7/(n+4) ; n =3 8/(n+4); n= 4 14/(n+4); n=3, 10 28/(n+4); n=3,10,24 56/(n+4); n=3,4,10,24,52

3+4+10+24+52 = 93

Solve by Microsoft Excel ^^