Question

please solve it please please please please please please yrr​

Answer 1

Let the digits at ones, tens and hundreds place be (a - d)a(a−d)a and (a + d)(a+d) respectively.

The, the number is

= (a + d)×100 + a ×10 + (a - d)

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

= 111a+99d........(1)

The number obtained by reversing the digits is

= (a - d) ×100 + a×10 + (a + d)

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

= 111a−99d

It is given that the sum of the digits is 15.

(a - d) + a + (a + d) =15

3a = 15

a = 5

Also it is given that the number obtained by reversing the digits is 594 less than the original number.

111a - 99d = 111a + 99d - 594

−99d = 99d−594

-198d = -594

d = 3

From (1)

So, the number original number is

111(5) + 99 (3) = 852.