Question

Let $x,y,z,t$ are in A.P. and ${1000}^{{\log }_{10}x}+{1000}^{{\log }_{10}y}+{1000}^{{\log }_{10}z}={1000}^{{\log }_{10}t}$. If $x,y,z,t$ are positive integers, then the minimum possible value of $x+y+z+t$ is

• A. $18$
• B. $15$
• C. $12$
• D. $21$

Answer 1

The correct option is A $18$

Let $x=a,y=a+d,z=a+2d,t=a+3d$
Now, ${1000}^{{\log }_{10}x}+{1000}^{{\log }_{10}y}+{1000}^{{\log }_{10}z}={1000}^{{\log }_{10}t}$
$\Rightarrow x^3+y^3+z^3=t^3\Rightarrow a^3+(a+d{)}^3+(a+2d{)}^3=(a+3d{)}^3\Rightarrow 2a^3-12a{d}^2-18d^3=0\Rightarrow {\left(\frac{a}{d}\right)}^3-6\left(\frac{a}{d}\right)-9=0$
Assuming $\frac{a}{d}=b$
$\Rightarrow b^3-6b-9=0$
By observation, $b=3$ is a root of the equation.
$\Rightarrow (b-3)(b^2+3b+3)=0\Rightarrow b=3(\because b^2+3b+3>0)$

$\frac{a}{d}=3\Rightarrow a=3d$
$\therefore x=3d,y=4d,z=5d,t=6d\Rightarrow x+y+z+t=18d$
As $x,y,z,t$ are positive integers, so the minimum value of $x+y+z+t$ is $18$.