Question

Find a three-digit number if its digits form a geometric progression and the digits of the new three digit number which is smaller by 400 form an arithmetic progression.

Answer 1

$\text { Let } N=x y z \Rightarrow y^2=x z $
$\text { number smaller by } 400 \text { is }(100 x+10 y+z)-400$
$\text { digits are }(x-4), y, z (x \geq 4) $
$\text { given } x-4, y, z \text { are in A.P. } \Rightarrow 2 y=x-4+z $
$ 2 \sqrt{x z}=x+z-4 $
or $ x+z-2 \sqrt{x z}=4 $
$(\sqrt{x}-\sqrt{z})^2=4$
$ \sqrt{x}-\sqrt{z}=2 \text { or }-2 \Rightarrow x \text { and } z \text { must be a perfect square } $
$\text { but } x>4 \Rightarrow x=9 $
$\therefore 3-\sqrt{z}=2 \Rightarrow z=1 $
$\text { or } 3+\sqrt{z}=2 \Rightarrow \sqrt{z}=-1 \text { (not possible) } $
$\text { If } x=9, z=1 \Rightarrow y=3$
$\text { Hence } N=931$