The digit of a positive number …

Let the digits of the required 3 - digit number at hundreds, tens and ones places be  {tex}a - d, ~ a , ~ a + d{/tex} respectively
Then their sum = 15
i.e, {tex}a - d + a + a + d{/tex} = 15
{tex}\Rightarrow{/tex} {tex}3a{/tex} = 15
 {tex}\Rightarrow{/tex} {tex}a{/tex} = 5
Required three digit number = {tex}100(a - d) + 10a + a + d{/tex}
= {tex}100a - 100d + 10a + a + d{/tex}
= {tex}111a - 99d{/tex}
Number obtained by reversing the digits = {tex}100(a + d) + 10a + a - d{/tex}
= {tex}100a + 100d + 10a + a - d{/tex}
= {tex}111a + 99d{/tex}
According to the question,
{tex}111a + 99d = 111a - 99d - 594{/tex}
{tex}\Rightarrow{/tex} {tex}594 = 111a - 99d - 111a - 99d{/tex}
{tex}\Rightarrow{/tex}{tex}594 = -198d{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { - 594 } { 198 } = d{/tex}
{tex}d = -3{/tex}
The number is 111a - 99d
{tex}111 \times 5 - 99 \times - 3{/tex}= 852
555 + 297 = 852
Therefore, Number is 852.