​The digits of a three digit natural number are in A.P and their sum is 18. The number obtained by reversing the digits is 594 less than the original number . Find the original number . plz ans it fast . give me a solution . its urgent
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Dear Student,

Please find below the solution to the asked query:

Given : ​The digits of a three digit natural number are in A.P and their sum is 18.

Let our three digits are :  a -  d  ,  a  and a  + d , So 

a -  d +  a  + a  + d  =  18 

3 a =  18 

a  = 6 

Also given : The number obtained by reversing the digits is 594 less than the original number .

Our number :  100 ( a - d ) + 10 a  + 1 ( a + d ) = 100 a  - 100 d +  10 a +  a  +  d =  111 a - 99 d

Number we get after reversing digits : 100 ( a + d ) + 10 a  + 1 ( a - d ) = 100 a + 100 d +  10 a +  a  -  d =  111 a + 99 d

Then from given condition we get

111 a -  99 d =  111 a +  99 d +  594

- 198 d =  594

d = - 3

So,

a  -  d  =   6   - ( - 3 ) =  6 + 3  =   9 ,  a  = 6   and  a  +  d  =   6   +   ( - 3 ) = 6 - 3 = 3

Theerfore,

Our three digit number  =  963                                                                                   ( Ans )

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