Least Value Of $x+y+z$

Question

If $x$, $y$ and $z$ are positive integer and $3x=4y=7z$, then What is the least possible value of $x+y+z$?

Answer 1

Since x, y and z are integers and positive, and 3, 4 and 7 are coprime, x must contain 4 and 7 as factors (that is, the minimum value for x is 28), y must contain 3 and 7 (21) and z must contain 3 and 4 (12) Then 3*(4*7*n)=4*(3*7*m)=7*(4*3*p), where m,n and p are positive integers So the least possible value for x+y+z is when n=m=p=1 and x+y+z=28+21+12

Answer 2

Hint : Think of 3x , 4y and 7z as surfaces in x y z plane and your are trying to find out the the common point of intersection of these surfaces . If there are more than one you select the least

Answer 3

Let $f(x,y,z)$=$x+y+z$
By given relations,
$y=3/4x$
$z=3/7x$
As $x$ , $y$ , and $z$ are positive integers,
Therefore, $x$ is a multiple of $28$
Now, $f(x,y,z)$=$61/28x$
For min$f(x,y,z)$,
Clearly, $x=28$ and min value is $61$