I have decided to study math on my own and I came across these questions. Would really appreciate if anyone could help and explain how to solve these.

Question

20.) How many terms of the sequence with general term $a_{n}=\dfrac{3n-72}{n}$ are integers?

21.) How many terms of the sequence with general term $a_{n}=\dfrac{n^{3}+4n^{2}+3n+1}{n+2}$ are integers?

I could not come up with a solution for both of the problems, so help is appreciated

Answer 1

For your first question, notice the following:

$$\text{a}_\text{n}:=\frac{3\text{n}-72}{\text{n}}=\frac{3\text{n}}{\text{n}}-\frac{72}{\text{n}}=3-\frac{72}{\text{n}}\tag1$$

Finding the divisors of $72$ gives:

$$\left\{1,2,3,4,6,8,9,12,18,24,36,72\right\}\tag2$$

Answer 2

20. we must have $n\mid 72$, which can be listed by brute force.
21. Do division algorithm and find that $n+2\mid 3$, and continue from here similarly.

Answer 3

For the first question, you can rewrite it like this:

$$\frac{3n-72}{n}=\frac{3n}{n}-\frac{72}{n}=3-\frac{72}{n}$$ This means that $\frac{3n-72}{n}$ will only be an integer if $n$ is a factor of $72$ (it may be a positive or negative factor of $72$).

For the second question, try using algebraic long division and use a similar method to the one I've employed in the first question.

If you need any more help, please ask :)

Answer 4

For the second question,

$$n^3 + 4n^2+3n+1 \equiv (-2)^3+4(-2)^2+3(-2)+1\equiv 3 \pmod{n+2}$$

so $(n+2) | 3, n+2=1, -1, 3, -3$.

(I assume $n$ is an integer.)

If you don't know modular math, then

$$n^3+4n^2+3n+1=n^3+2n^2+2n^2+4n-n+1=(n+2)(n^2+2n-1)+3$$

Again $\frac{3}{n+2}$ is an integer.