Question

The digits of a positive number of 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 three digits of a positive number be

$a-d,a,a+d$

$\therefore a-d+a+a+d=3a=15$

$\Rightarrow a=5$

Original number = $100(a-d)+10\left(a\right)+a+d$

= $100a-100d+10a+a+d$

= $111a-99d$

And, number obtained by reversing the digits

= $100(a+d)+10\left(a\right)+a-d$

= $100a+100d+10a+a-d$

= $111a+99d$

According to the given condition, $(111a-99d)-(111a+99d)=594$

$-198d=594$

$d=-3$

$\therefore$ Original number is $111\left(5\right)-99(-3)$

i.e., 852