Question

Given a three digit number whose digits are three successive terms of a G.P. If we subtract 792 from it, we get a number written by the same digits in the reverse order. Now if we subtract four from the hundred's digit of the initial number and leave the other digits unchanged, we get a number whose digits are successive terms of an A.P. Find the number.

Answer 1

Answer: 931

Let the number be $100x+10y+z$
$y^2=xz$ ----(1)
$100x+10y+z-792=100z+10y+x$
$\text{ or }100(x-z)+(z-x)=792\Rightarrow x-z=\frac{792}{99}=8$
Now, $x-4,y,z$ are in A.P.
Hence $2y=x-4+z$
$\Rightarrow y=\frac{x+z-4}{2}$ ----(2)
$\Rightarrow (x+z-4{)}^2=4\times z$
$(x+z{)}^2-8(x+z)+16=4xz$
$\Rightarrow (x-z{)}^2-8(x+z)+16=0$
$\Rightarrow 64-8(x+z)+16=0\Rightarrow x+z=10\dots (3)$
On solving $x=9,y=3\&z=1\Rightarrow N=931$