Question

If $4x+3y=121$, find how many positive integer solutions are possible?

• A. $10$
• B. $1$
• C. $0$
• D. Cannot be determined

Answer 1

The correct option is A $10$

We have to find only non-negative integer solution.

Coefficient of $x$ and $yare4$ and $3;$ The $L.C.M$ is $6.$

We can consider a packet of $12$ which may contain three $4s$ or four $3s$ but not both.

In $121$ we have such $10$ packets with $1$ remainder. If we include a zero packet it will be $11$.

However, the remainder 1 remains isolated and can not be broken into $4s$ and $3s$. So we have to unpack $1$ packet and add the remainder $1$ to make it $13$. This can be distributed as one $4$ and three $3s.$ So we have in all $10$ packets to distribute in $4s$ and $3s$.

Hence, the number of positive integer solutions are $10$.