Question

You have a total supply of $1000$ pieces of candy, and an empty vat. You also have a machine that can add exactly $5$ pieces of candy per scoop to the vat, and another machine that can remove exactly $3$ pieces of candy with a different scoop from the vat. When these two machines are done, there is only one piece of candy left in the vat. What is the smallest possible number of times the the first machine added candy to the vat?

Answer 1

Let $x$ be the number of times the first machine added candy and $y$ be the number of times the second machine removed candy.

From the problem, we have the equation: $5x - 3y = 1000 - 1 = 999$.

To minimize $x$, we want to make $y$ as large as possible. Since the second machine removes $3$ candies each time, $y$ must be a multiple of $3$. The smallest multiple of $3$ that is greater than $999$ is $1002$.

Therefore, $5x - 3(1002) = 999 \implies 5x - 3006 = 999 \implies 5x = 4005 \implies x = \boxed{801}$.

Therefore, the smallest possible number of times the first machine added candy to the vat is $801$.