+102 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
+128578 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
+26367 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
\(\begin{array}{|rclcl|} \hline ax &=& by &=& cz\\ \hline \end{array}\\ \begin{array}{rclcl} ax &=& by && & \qquad \Rightarrow & y=\frac{a}{b}x\\ &&by &=& cz && \\ ax & & &=& cz &\qquad \Rightarrow & z=\frac{a}{c}x \end{array}\)
\(\begin{array}{|rclcl|} \hline \not{x} +y+z &=& \not{x} + \frac{a}{b}x + \frac{a}{c}x \\ y+z &=& \frac{a}{b}x + \frac{a}{c}x \\ y+z &=& x \cdot \left( \frac{a}{b}+ \frac{a}{c} \right) \\ y+z &=& x \cdot \left( \frac{ac+ab}{bc} \right) \\ y+z &=& x \cdot \frac{(ac+ab)}{bc} \qquad & | \qquad x_{\text{the least possible value}}=bc\\ y+z &=& bc \cdot \frac{(ac+ab)}{bc} \\ y+z &=& \frac{bc}{bc}\cdot (ac+ab) \\ y+z &=& ac+ab \\ y+z &=& \underbrace{ac}_{=y}+\underbrace{ab}_{=z} \\\\ x &=& bc \\ y &=& ac \\ z &=& ab \\ x+y+z &=& bc+ac+ab \\ \hline \end{array}\)
\(\begin{array}{|rclcl|} \hline 3x=4y=7y && \qquad | \qquad a=3 \qquad b = 4 \qquad c = 7 \\ x+y+z &=& bc+ac+ab \\ x+y+z &=& 4\cdot 7 + 3\cdot 7 + 3\cdot 4 \\ x+y+z &=& 28 + 21 + 12 \\ x+y+z &=& \underbrace{28}_{=x} + \underbrace{21}_{=y} + \underbrace{12}_{=z} \\ x+y+z &=& 61 \\ \hline \end{array} \)
