integers check?

Question

HI,
Find all the possible values fir the integer "x" so that 5 / x-5 is also an integer.
i tried doing it and i got: 0, 6, and 10
can someone please help me if there are any more?
thanks in advance!

Answer 1

In order for 5 / x - 5 to be an integer, x -5 must go evenly into 5.

The only integers that go evenly into 5 are -5, -1, 1, and 5.

So we must have one of the following:
x - 5 = -5
x - 5 = -1
x - 5 = 1
x - 5 = 5

Solving these, we get:
x = 0
x = 4
x = 6
x = 10.

So these are the only solutions.

Answer 2

For this to have any possibility of being an integer then
-5 < x - 5 < 5 (with equality as well but I can't do that symbol) as anything outside this range would produce a fraction. As you are told that x is an integer this gives only a limited number of values to check and gives your results and x = 4.

Answer 3

assume 5/(x-5) = N and get x to 1 side. so x= 5/N +5. This is an integer if 5/N is an integer. since 5 is prime this is simple 5/N is an integer when N = 1,5,-1,-5. Therefore u missed one. Simply subsituting all these numbers for N gives u the solutions, the one u missed is when x = 4, since -5 is an integer.

Answer 4

An integer can be positive or negative. Thus, 0, 4, 6, and 10 are correct. Any negative values of x would simply not work, so those are the only answers (that I can think of).

Answer 5

What about 4?

Answer 6

5/(x-5) = n say
or 5 = nx -5n
or 5(n+1) = nx
x = 5(n+1)/n
for x to be integer

n should devide 5 that is -5,-1,1,and 5
evaluate x for each n
n = -5 , x = 4
n = -1 x = 0
n = 1 x = 10
n = 5 x = 6

so theser are solutions

Answer 7

(-2)(-9)