Question

The digits of a positive integer, having three digits are in A.P. and their sum is $15$. The number obtained by reversing the digits is $594$ less than the original number. Find the number.

Answer 1

Let the digits at ones, tens and hundreds place be $(a-d)a$ and $(a+d)$ respectively. The, the number is
$(a+d)\times 100+a\times 10+(a-d)=111a+99d$
The number obtained by reversing the digits is
$(a-d)\times +a\times 10+(a+d)=111a-99d$
It is given that the sum of the digits is $15$.
$(a-d)+a+(a+d)=15$ ...(i)
Also it is given that the number obtained by reversing the digits is $594$ less than the original number.
$\therefore 111a-99d=111a+99d-594$ ...(ii)

$⟹3a=15$ and $198d=594$
$⟹a=5$ and $d=3$

So, the number is $111\times 5+99\times 3=852$.