Three of them

$\begin{cases} x + 2y & = p + 6 \\ 2x - y & = 25 - 2p \end{cases}$

Let $(x,y)$ be positive real integers that satisfy the system of equations above. How many possible value of $p$ ? And for what value of $p$ ?

Easy Math Editor

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Comments

Solving for $x$ and $y$ will give:

$x=\dfrac{56-3p}5 \text{ and } y= \dfrac{4p-13}5....(1)$ $\because x,y>0\implies \dfrac{13}{4}< p<\dfrac{56}3....(2)$

Looking at original first equation: $x+2y-6=p$ we can conclude $p$ is also an integer which along with $(2)$ gives $p\in\{4,5,6,...,18\}$ .

From $1$ , now since $y$ is an integer it's numerator must be a multiple of $5$ and for that $4p$ must end in $8$ or $3$ so when $13$ is subtracted, the numerator ends in $5$ or $0$ respectively and hence divisible by $5$ . 4p cannot end in $3$ but ends in $8$ when $p\equiv 5n+2$ or $p\in\{7,12,17\}$ . For these values of $p$ , $x$ also assumes an integer.

$p=7,~~ (x,y)=(7,3)$ $p=12, ~~(x,y)=(4,7)$ $p=17,~~(x,y)=(1,11)$

PS: I might have missed some solutions bcoz I'm pretty good at it.

Rishabh Jain - 4 years ago

Nice solution! It's important to notice that $\dfrac{13}{4} < p < \dfrac{56}{3}$ . Thanks!

Fidel Simanjuntak - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$