Question

the digits of a three digit number are in AP their sum is 18 and the number obtained by reversing the digits is 594 less than the original number find the original number​

Answer 1

Step-by-step explanation:

Let the hundred's, ten's and unit digit's be a+d,a,a−d

Then, a+d+a+a−d=15

⇒3a=15⇒a=5

[100(a+d)+10a+(a−d)]−[100(a−d)+10a+(a+d)]=594

⇒99d+99d=594⇒198d=594⇒d=3

Hence, the requried number is 852