# Math Help - Finding integers? Point of tangency?

1. ## Finding integers? Point of tangency?

Intro to Calc HELP!?????? Finding integrs?

1)Find an integer, x, greater than 4 where the sum of x consecutive integers is divisible by x

2) Show that the equation above gives the correct slope for the line tangent to a circle with a radius of 13 at the point (-5,12)

So I'm thinking, the circle's center is at (0,0) because then the point of tangency will be (-5,12) exactly...But I don't know what to do

For #1, I got 5n+10....is that right or wrong?

2. Assuming the circle is centered at the origin, then its equation above the x-axis is $y=\sqrt{169-x^{2}}$

We can use this and its derivative to find the slope at (-5,12) to find the equation.

$y'=\frac{-x}{\sqrt{169-x^{2}}}$

Using y=mx+b, we have $12=\frac{5}{12}(-5)+b$

$b=\frac{169}{12}$

$y=\frac{5}{12}x+\frac{169}{12}$ is the line equation.

3. Ok now im lost, how do i do #1 then?

I got 5n+10..i dont know how the two relate now.....

4. You are correct, $X=5$.
Note the sum is $\sum\limits_{k = 0}^4 {\left( {5N + k} \right)} = 5N + \frac{{4(5)}}{2} = 5N + 10$ which is clearly divisible by 5.

5. Hello, realintegerz!

Could you state the original problem?
I'm sure that this isn't what "they" said.

1) Find an integer $n > 4$ where the sum of $n$ consecutive integers
is divisible by $n$.
Let $a$ be the first (smallest) integer.

$\text{Then: }\;S \;=\; \underbrace{a + (a + 1) + (a + 2) + (a + 3) + \hdots + (a + [n-1])}_{n\text{ terms}}$

. . . . $S \;=\;\underbrace{(a + a + a + \hdots + a)}_{n\text{ terms}} + (1 + 2 + 3 + \hdots + [n-1])$

. . . . $S\;=\;na + \frac{n(n-1)}{2} \;=\;n\left(a + \frac{n-1}{2}\right)$

$S$ is divisible by $n$ if $\left(a + \frac{n-1}{2}\right)$ is an integer.
. . This happens when $n$ is odd.

Therefore, $n$ is any odd integer greater than 4.

2) Show that the equation above gives the correct slope for the line tangent
to a circle with a radius of 13 at the point (-5,12)
I don't see what this question has to do with #1.
. . There is no "equation above."

I will assume that the circle is centered at the origin.

The slope of the radius from (0, 0) to (-5, 12) is: . $\frac{12-0}{-5-0} \:=\:-\frac{12}{5}$

The tangent at the point is perpendicular to that radius.

Therefore, the slope of the tangent is: . $\frac{5}{12}$

6. Well actually it's a series of 4 problems

1) Write an expression in terms of an integer, a, for the sum of any 3 consecutive integers and show that this sum is divisible by 3

a+(a+1)+(a+2)
=3a+3

3(a+1)

3(a+1)/3= (a+1) x 3/3 = (a+1) x 1 = a+1

2) Wrote an expression in terms of an integer, a, for the sum of any 4 consecutive integers and show that this sum isn't divisible by 4

a+(a+1)+(a+2)+(a+3)
=4a+6

2(2a+3)

2(2a+3)/4 = (2a+3) x 2/4 = (2a+3) x 1/2 = (2a+3)/2

3) Find an integer, n, greater than 4 where the sum of n consecutive integers is divisible by n

4) Show that the equation above gives the correct slope for the line tangent to a circle with radius of 13 at the point (-5,12)

Yeah im lost on how 4 relates to 3....should i just ignore it?