Question

How many solutions in positive integers are there of the equation $5x+7y$ $=$ $397$?

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is D 12

Begin by trying to find bounds for the values of $x$ or $y$. It is easier to find bounds for $y$. These are:
$0\le y\le \left(\right[397÷7]=56)$
Now, the value of $y$ must be such that $397-7y$ is divisible by five. For that, $7y$ must end in $2$ or $7$. Observing the pattern of the units digits' of the multiples of seven: $\{7,4,1,8,5,2,9,6,3,0\}$, the allowed values of $y$ come out to be:
$1,6,11,16,21,\mathrm{26....51},56$.
This is also the number of solutions, because every $y$ gives a unique $x$.