Integer Solutions

If we assume the expression to be $k$ , we have $n=\frac{6k+3\pm\sqrt{16k^2+96k+9}}{2k}$ . Since the roots must be integral, discriminant must be a perfect square, so $16k^2+96k+9=L^2\implies 135=(4k+12)^2-L^2=(4k+12-L)(4k+12+L)$ . We check all the possibilities namely $1\times 135, 3\times 45, 5\times 27$ , $9\times 15, -1\times -135, -3\times -45$ , $-5\times -27, -9\times -27$ and arrive at the solutions $\{-5,-1,0,2,3,4,7,10\}$ .

If $(n-1)(n-5) | (3n+15)$ , then $(n-1) | (3n+15)$ . Since $3n+15\equiv 18 \pmod {n-1}$ , thus $n-1\in\{-18,-9,-6,-3,-2,-1,1,2,3,6,9,18\}$ , which means $n\in \{-17,-8,-5,-2,-1,0,2,3,4,7,10,19\}$ . Checking all of these $n$ , we see that only $n=-5,-1,0,2,3,5,7,10$ works. Their sum is $20$ , as desired.

[Note: Similar solutions that checked for $(n-1) | (3n+15)$ and $(n-5) | (3n+15)$ , would also need to check that the intersection of the solution sets actually yields an integer value to $\frac {3n+15}{(n-1)(n-5)}$ .

If $\frac{3n+!5}{(n-1)(n-5)}$ is an integer, then (n-1)(n-5) is a factor of 3n+15, so either $3n+15=0$ or $(n-1)(n-5)\leq 3n+15$ . The first case yields $n=-5$ which is within the domain. The second case yields $(n-1)(n-5)\leq 3n+15$ $\implies n^2-6n+5\leq 3n+15$ $\implies n^2-9n-10\leq0$ $\implies (n-10)(n+1)\leq 0$ $\implies -1\leq n\leq 10.$ Thus the only $n$ that make the fraction an integer in the second case are $n=\{-1,0,2,3,4,7,10\}$ thus the sum of all these solutions is -5-1+0+2+3+4+7+10=20.

[Arithmetic Error corrected - Calvin]

First, we note that if the absolute value of the numerator is greater than that of the denominator, the fraction will never be an integer. Since the numerator is linear and the denominator is a quadratic with a positive coefficient for the quadratic term, there is clearly a range beyond which the numerator would be smaller in magnitude than the denominator.

*There is however one exception to this argument which is where the numerator is 0. Despite it being smaller than the denominator, the fraction is an integer. -- In this particular case, it is when $n = -5$ .

Denominator is larger in magnitude when:-

(a) If $3n+15 ≥ 0$ : $(n-1)(n-5) > 3n+15 \Rightarrow n^2 - 9n - 10 > 0 \Rightarrow ( n > 10$ or $n < -1 )$

(b) If $3n+15 < 0$ : $(n-1)(n-5) < -3n-15 \Rightarrow n^2 - 3n + 20 > 0$ , which is always true.

Hence we only need to consider the cases when $-1 ≤ n ≤ 10$ for the fraction to be an integer. By simple listing of all cases, we find that the fraction is an integer when $n = -1, 0, 2, 3, 4, 7, 10$ , as listed below, and hence the answer is $10+7+4+3+2-1-5=20$

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When $n = -1$ : Fraction = $\frac{3(4)}{(-2)(-6)}$ - Integer

When $n = 0$ : Fraction = $\frac{3(5)}{(-1)(-5)}$ - Integer

When $n = 1$ : Fraction = $\frac{3(6)}{(0)(-4)}$ - Undefined

When $n = 2$ : Fraction = $\frac{3(7)}{(1)(-3)}$ - Integer

When $n = 3$ : Fraction = $\frac{3(8)}{(2)(-2)}$ - Integer

When $n = 4$ : Fraction = $\frac{3(9)}{(3)(-1)}$ - Integer

When $n = 5$ : Fraction = $\frac{3(10)}{(4)(0)}$ - Undefined

When $n = 6$ : Fraction = $\frac{3(11)}{(5)(1)}$ - Noninteger

When $n = 7$ : Fraction = $\frac{3(12)}{(6)(2)}$ - Integer

When $n = 8$ : Fraction = $\frac{3(13)}{(7)(3)}$ - Noninteger

When $n = 9$ : Fraction = $\frac{3(14)}{(8)(4)}$ - Noninteger

When $n = 10$ : Fraction = $\frac{3(15)}{(9)(5)}$ - Integer

Note that if n<=6, the absolute value of the denominator is greater than the absolute value of the numerator, so there are no solutions in this interval. Using casework, you can find the solutions n =-5, n=-1, n=0, n=2, n=3, n=4, n=7, and n=10. If n>=11, the absolute value of the denominator is greater than the absolute value of the numerator, so there are no solutions in this interval. Therefore, the answer is 20.

Let $\frac {3n+15}{(n-1)(n-5)}$ =k, where k is an integer.

Cross-multiplying and rearranging,

We get k\((n^2)\ - (6k+3)n + (5k-15) =0

Solving the quadratic equation,

n= \(\frac{6k+3 \pm \sqrt{(4k+12)^2-135}}{2k} ---(1)

For n to be an integer,

\((4k+12)^2-135 = a^2\) for some positive integer a.

$(4k+12)^2-a^2$ = 135

(4k+12+a)(4k+12-a) =135

Note that 135= $3^3$ x 5(Let pm represent +-)

135 = $\pm 1$ x $\pm 135$

   = \\\(\\pm 3\\\) x \\\(\\pm 45\\\)

   = \\\(\\pm 5\\\) x \\\(\\pm 27\\\)

   = \\\(\\pm 9\\\) x \\\(\\pm 15\\\)

Since 4k+12+a and 4k +12-a are both integers.

Testing all 8 possibilities ,

     Note : Column 1:4k+12+a

          Column 2:4k+12-a

           Column 3: a

              Column 4:k

              Column 5:n

   135.              1.          67.    14.       -

     45.              3.          21.      3.      0,7

     27.              5.          11.      1.    -1,10

     15.              9.           3.       0.       -5

     -9.              -15          3.     -6.        3

     -5.              -27.         11.    -7.       2

     -3.              -45.         21.    -9.       4

     -1.              -135.       67.    -20.     -

Therefore, the answer is 0+7-1+10-5+3+2+4 = 20.

We note that since n is an integer, n-1 and n-5 are therefore integers.

Therefore, (3n+15)/(n-1) and (3n+15)/(n-5) are integers.

By applying a concept similar to Euclidean Algorithm, we get that:

18/(n-1) and 30/(n-5) are integers.

Therefore, the possible values of n are taken from the intersection of the two sets {x:(x-1)|18} and {x:(x-5)|30}.

Or

{-17, -8, -5, -2, -1, 0, 2, 3, 4, 7, 10, 19} and {-25, -10, -5, -1, 0, 2, 3, 4, 6, 7, 8, 10, 11, 15, 20, 35}.

This gives possible values of x as in the set {-5, -1, 0, 2, 3, 4, 7, 10}. Checking each value individually, we note that all work, giving -5-1+0+2+3+4+7+10=20.

Note: The two abovementioned conditions being satisfied individually is not a sufficient condition for the expression to be an integer.

A fraction is an integer if the numerator is 0 and the denominator is not equal to 0 or if the numerator is a multiple of the denominator. Since the denominator can’t equal 0 we have $n \neq 1$ and $n \neq 5$ .

Case 1: Numerator is 0. This occurs when $3n+15=0 \Rightarrow n=-5$ . The denominator is non-zero, so this is a valid solution.

Case 2: Numerator is a multiple of the denominator. A necessary condition for this to occur is $| 3n+15 | \geq |n^2-6n+5|$ . Squaring both sides and rearranging, we have $\begin{aligned} 0 & \geq (n^2-6n+5)^2 - (3n + 15)^2 & \\ & \geq (n^2-9n-10)(n^2-3n+20) & \\ & \geq (n-10)(n+1)(n^2-3n+20) &\\ \end{aligned}$

Since $n^2-3n+20 = (n-1.5)^2 + 22.5$ , it is always positive. So $-1 \leq n \leq 10$ are the only possible solutions. Checking these cases, we get that $\frac {3n+15}{(n-1)(n-5)}$ is an integer for $n =-1, 0, 2, 3, 4, 7$ and $10$ .

Hence the sum of all $n$ is $-5 -1 + 0 + 2 + 3 + 4 + 7 + 10 = 20$ .

Since 3n+15/(n-1)(n-5) is an integer (n-1)(n-5) divides 3n+15 therefore n-1 divides 3n+15 i.e. n-1 divides 3(n-1)+18 we know (n-1) divides 3(n-1) therefore (n-1) divides 18 therefore possible values of (n-1) are 1,2,3,6,9,18,-1,-2,-3,-6,-9,-18 therefore (n-5) = (n-1)-4 = -3,-2,-1,2,5,14,-5,-6,-7,-10,-13,-22

          Again (n-5) divides 3n+15 
                  therefore (n-5) divides 3(n-5)+30
                  since (n-5) divides 3(n-5)
                              (n-5) divides 30
           therefore among the possible values of (n-5) the divisors of 30 only -3,-2,-1,2,5,-5,-6,-10 divide 30

others such as 14,-7,-13,-22 do not divide 30 therefore the values of (n-5) are -3,-2,-1,2,5,-5,-6,-10 therefore the values of n are (-3+5)=2 (-2+5)=3 (-1+5)=4 (2+5)=7 (5+5)=10 (-5+5)=0 (-6+5)=-1 (-10+5)=-5 sum of all possible values of n =2+3+4+7+10+0+(-1)+(-5) =20 ANS : 20

Firstly, (n-1) and (n-5) must be divisors of (3n +15).

(1) We can express (3n + 15) as (3n- 3) + 18. Clearly 18 must be divisible by (n - 1).

(n-1) can take the value of 18, 9, 6, 3, 2, 1, -1, -2, -3, -6, -9, -18.

n can take the value of 19, 10, 7, 4, 3, 2, 0, -1, -2, -5, -8, -17.

(2) We can express (3n + 15) as (3n -15) + 30. Clearly 30 must be divisible by (n-5).

(n-5) can take the value of 30, 15, 10, 6, 5, 3, 2, 1, -1, -2, -3, -5, -6, -10, -15, -30.

n can take the value of 35, 20, 15, 11, 10, 8, 7, 6, 4, 3, 2, 0, -1, -5, -10, -25.

Comparing the possible values of n that satisfy both conditions, we get n= 10 or 7 or 4 or 3 or 2 or 0 or -1 or -5.

The required sum is 20.

Here are the possible integer values: 10, 7, 4, 3, 2, 0, -5, -1. The sum is 20.

when n more than 10 it's impossible doesn't have decimal or n smaller than -2 except -5. so just test 2,3,4,6,7 and 6 is incorrect. answer is 2+3+4+7+10-6=20