Question

the digits of a positive number of three digits are in A.P. sum is 15 . the number obtained by reversing the digit is.594 less than the original number . find the numbers .​

Answer 1

let the original three-digit number be xyz .

since digits x,y and z are in AP (given)

therefore,

let x = a - d 

    y = a

    z = a + d

(where a and d are the first term and common difference respectively.)

now, according to the question,

x + y + z = 15

a - d +a + a + d = 15

3a = 15

a = 5

So , a = 5 ----- ( 1 )

Also, another condition given is

zyx = xyz - 594

100.z + 10.y + x = 100.x + 10.y + z - 594

100( a + d ) + 10( a ) + ( a - d ) = 100( a - d ) + 10( a ) + ( a + d ) - 594

on further solving the above equation, we get

d = - 3 ----- (2)

NOW,

x = a - d = 5 - (-3) = 8                           (from 1 and 2)

y = a = 5                                               (from 1)

z = a + d = 5 + (-3) = 2                          (from 1 and 2)

Thus, the required number is 852.

Hope this helps!