Positive integers $x$,$y$,$z$

Question

If $x$ , $y$ and $z$ are positive integers and $3x = 4y = 7z$, then calculate the smallest possible value for $x+y+z$.

How do you do this? Can someone please give me a hint?

Answer 1

Consider $N = 3x = 4y = 7z$. $N$ is known to have divisors $3$, $4$ and $7$. The smallest positive $N$ that has these $3$ divisors is $84$.

And since $N, x, y, z$ and $x+y+z$ are all related by constant ratios, the smallest $N$ gives the smallest $x+y+z$.

Answer 2

$k=3x=4y=7z\implies x=\dfrac{k}{3},y=\dfrac{k}{4},z=\dfrac{k}{7}\implies x+y+z=k\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{7}\right)=k\left(\dfrac{4\cdot7+7\cdot3+3\cdot4}{84}\right)\geq \dfrac{61}{84}$

Answer 3

I know that is a simple solution but, with your condition the smallest is $0$ with $x=y=z=0$ .

Answer 4

Since each common multiple of $3$, $4$, and $7$ is a multiple of their least common multiple, the smallest sum $x + y + z$ is found by finding the $x$, $y$, and $z$ such that $3x = 4y = 7z$ is equal to the least common multiple of $3$, $4$, and $7$.

Answer 5

Let $S=x+y+z$.

$$S=x+y+z=x+\frac{3x}4+\frac{3x}7$$

But the three terms are integer, so $x$ must be a multiple of $7$ and $4$, that is, $x=28k$ with $k$ positive integer, so we get: $$S=28k+21k+12k=61k$$ Therefore, the least possible value for $S$ is $61$.

Answer 6

One way to approach the problem would be to restate it as an Integer Programming (IP) Problem, and solve it as an Integer Programming (IP) problem.