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

Answer: 931

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