How to find the cost of a tv set when it is given as an algebraic expression? The problem is as follows:
Gabriel wants to purchase a new TV for her mother as a Christmas gift. The cost is $2P$ thousand dollars. In order to fulfill this wish he begins saving his earnings as a pharmacist clerk for six months. Assume that the money he earns is the same each month and $P$ to be the sum of the coefficients of the rational integer algebraic expression and not null of the form
$$M(x,y)=\left(\frac{n+4}{3}\right)x^{5-n}y^{4-m}-2x^{n-3}y^{m-4}+(n-3)y^{n+2}$$
If Gabriel decides to buy the gift after $4$ months instead. By how much additional dollars should Gabriel have to save each month with respect of his original plan so that he can buy the tv?
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{300 dollars}\\
2.&\textrm{400 dollars}\\
3.&\textrm{500 dollars}\\
4.&\textrm{200 dollars}\\
\end{array}$
How exactly should be the relationship between the given condition of the integer rational integer algebraic expression should be used?. Can someone help me here?.
So far what I've been able to spot was this:
$P=\frac{n+4}{3}-2+(n-3)$
$P=\frac{4n-11}{3}$
But that's the part where I'm stuck, where exactly should I go from here?. I don't know what else to do with the information related with those $4$ months.
 A: Say he saves $s$ for $6$ months. Then $$6s = 2000P \implies s=\frac{1000}{3} P$$
Say he saves $s’$ for $4$ months. Then $$4s’=2000P \implies s’=500P$$
Then $$s’-s = \frac{500}{3} P = 500 \times \frac{4n-11}{9} $$ Now, this must be equal to exactly one of $200,300,400,500$. Set $s’-s = 100a$ with $a\in\{2,3,4,5\}$ so that $$5(4n-11) = 9a \implies n= \frac{9a+55}{20} $$ $n$ must be an integer, so $20$ must divide $9a+55$. That means $9a+55$ must end in a $0$, and that only happens for $a=5$ and so $$s’-s =100\cdot 5 =500$$
A: Note that rational integer algebraic expression means that the coefficients are integers, the exponents are rational and the expression is rational of the form $\frac{P(x,y)}{Q(x,y)}$. Also note that $n$ is in the exponent of the variables $x$ and $y$.
Thus $n$ is rational, more so $n$ must be an integer because we have the coefficient $n-3$, furthermore $\frac{n+4}{3}$ must be an integer, and  $3$ divides $n+4$.
Because $\frac{n+4}{3}$ must be an integer, so must $P$, and since Gabriel saves an equal amount each month he must save either $M$ or $M’$ dollars every month respectively:
$$6M=2000P$$
$$4M’=2000P$$
Since $P$ is an integer so are $M$ and $M’$.
$$M’-M=2000P \cdot \left(\frac{1}{4} - \frac{1}{6}\right)=\frac{500P}{3}$$
Since $M’-M$ is an integer because $M’$ and $M$ are both integers $\frac{P}{3}$ must be an integer.
Given $P=\frac{4n-11}{3}$, $9$ divides $4n-11$ thus $n=9n'+5$ for arbitrary $n' \in \mathbb{Z}$. Thus $\frac{P}{3}=4n’+1$
$$M’-M=500\cdot (4n’+1)$$
Could we restrict $n’$ further given the fact that the algebraic expression is rational and $n$ appears in the exponent? Every polynomial is a rational function, thus there are no further restrictions on $n$, and therefore $n’$ as well.
From there, the only available answer that is a multiple of $500$ is $(3).$
