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We have the relationships :
$5x+6y=150\;\;,\;\;\;5x=150-6y\;\;\;6y=150-5x \tag{1}$
From $6\mid 150\;\;and\;\;6\mid 6y$ and from the second formula (1) it follows that $6\mid 5x$ and because $gcd(5,6)=1$ we have $6\mid x$ so
$x=6\cdot x_1\;,\;\;x_1\in\mathbb{N^*} \tag{2}$
Similarly, we have $5\mid 150\;\;and\;\;5\mid 5x$ , and from the third formula (1) it results $5\mid 6y$ , so
$y=5\cdot y_1\;\;\;y_1\in\mathbb{N^*} \tag{3}$
Then $5x+6y=150\;\;\overset{(2)}{\underset{(3)}{\Leftrightarrow}}\;\;\;5\cdot 6x_1+6\cdot 5y_1=150\Leftrightarrow$
$\Leftrightarrow\;\;\;\;\;\;\;\;\;\;x_1+y_1=5 \tag{4}$
If $x_1=n\in\mathbb{N^*}$ , then $y_1\overset{(4)}{=}5-x_1=5-n$ , so the nonzero natural number solutions for (4) are :
$x_1=n\;\;,\;\;\;y_1=5-n\;\;\;\;\;1\leqslant n \leqslant 4 \tag{5}$
But
$x+y\;\;\;\overset{(2)\;(3)}{\underset{(5)}{=}}\;\;6\cdot n+5\cdot (5-n)=n+25 \tag{6}$
and then $1\leqslant n \leqslant 4 \Leftrightarrow26\leqslant n+25 \leqslant 29\underset{(6)}{\Leftrightarrow}$
$\Leftrightarrow\;\;\;\;\;26 \leqslant x+y \leqslant 29 \tag{7}$
We have in (7) $x+y=26$ for $n=1$ , so $x_1=1\;,\;y_1=4$ and we get from (2) and (3) $x=6\;,\;y=20$ for the minimum value. And $x+y=29$ for $n=4$ , so $x_1=4\;,\;y_1=1$ so $x=24\;,\;y=5$ for the maximum value.
Since there are a small number of values for $x\; and\; y$ , we will list all the products $x\cdot y$ in the table below, from where we will also obtain the answer.
\begin{array}{c|c|c|c|c}n&1&2&3&4\\\hline x&6&12&18&24\\\hline y&20&15&10&5\\\hline x\cdot y&120&180&180&120\\\end{array}
$\blacksquare$
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