Value

3x = 4y = 7z   implies that

y = (3/4)x        and z  = (3/7)x

So....x + y + z  =

x + (3/4)x + (3/7)x      

And, if y and z are integers, x must be divisible by both 4 and 7.....so, the least that x  can be is 28

And  y = (3/4)(28)  = 21

And z = (3/7)(28)  = 12

So, the least possible value for x + y + z  =   28 + 21 + 12 =   61

If x,yand z are positive intergers and 3x = 4y = 7z then the least possible value of x+y+z is
A) 33
B) 40
C) 49
D) 61
E) 84

$\begin{array}{|rclcl|} \hline ax &=& by &=& cz\\ \hline \end{array}\\ \begin{array}{rclcl} ax &=& by && & \qquad \Rightarrow & y=\frac{a}{b}x\\ &&by &=& cz && \\ ax & & &=& cz &\qquad \Rightarrow & z=\frac{a}{c}x \end{array}$

$\begin{array}{|rclcl|} \hline \not{x} +y+z &=& \not{x} + \frac{a}{b}x + \frac{a}{c}x \\ y+z &=& \frac{a}{b}x + \frac{a}{c}x \\ y+z &=& x \cdot \left( \frac{a}{b}+ \frac{a}{c} \right) \\ y+z &=& x \cdot \left( \frac{ac+ab}{bc} \right) \\ y+z &=& x \cdot \frac{(ac+ab)}{bc} \qquad & | \qquad x_{\text{the least possible value}}=bc\\ y+z &=& bc \cdot \frac{(ac+ab)}{bc} \\ y+z &=& \frac{bc}{bc}\cdot (ac+ab) \\ y+z &=& ac+ab \\ y+z &=& \underbrace{ac}_{=y}+\underbrace{ab}_{=z} \\\\ x &=& bc \\ y &=& ac \\ z &=& ab \\ x+y+z &=& bc+ac+ab \\ \hline \end{array}$

$\begin{array}{|rclcl|} \hline 3x=4y=7y && \qquad | \qquad a=3 \qquad b = 4 \qquad c = 7 \\ x+y+z &=& bc+ac+ab \\ x+y+z &=& 4\cdot 7 + 3\cdot 7 + 3\cdot 4 \\ x+y+z &=& 28 + 21 + 12 \\ x+y+z &=& \underbrace{28}_{=x} + \underbrace{21}_{=y} + \underbrace{12}_{=z} \\ x+y+z &=& 61 \\ \hline \end{array}$

LCM(3,4,7) = 84

84/3 = 28          84/4 = 21         84/7 = 12 

3*28 = 4*21 = 7*12

28 + 21 + 12 = 61

.