Question

The digits of a the number of three digits are in A P and their sum is 15 the number obtained by reversing the digits is 594less than the original number find numbers

Answer 1

Let,the digit at the hundredth place of the number be (a+b).
Digit at the tens place be a.
Digit at the ones place be (a-d).
Sum of the digits = 15
=> (a+d) + a + (a-d) = 15
=> 3a = 15
=> a = 15/3
=> a = 5

The number formed by the digits
= 100 (5+d) + 10 (5) + (5-d)
= 555 + 99d

The number formed by reversing the digits
= 100 (5-d) + 10 (5) + (5+d)
= 555 - 99d

Given that the number formed by reversing the digits is 594 less

The original number is
(555+99d) - (555-99d) = 594
=> 198d = 595
=> d = 594/198
=> d = 3

So,
(a+d) = (5+3) = 8 ;
a = 5;
(a-d) = (5-3) = 2.

Hence the number formed by the digits

100×(8)+10×(5)+1×(2)
= 800+50+2
= 852