Question

The digits of a positive integer
whose three digits are in AP and
the sum is 21 The number
obtained by reversing the digits is
396 less than the original number
Find the number please​

Answer 1

Let 100x, 10y and z be the digits in the hundred's, ten's and unit places of the three digit number.

Then, by question z,10y and 109x are in AP

Now,100x+10y+z=21

or 3/2[2z+(3-1)d]=21

or 6z+6d=42

$s = n \div 2 \times {a + (n - 1)d} where \: n = number \: of \: terms \: a = first \: term \: and \: d = common \: difference$

or 6(z+d)=21

or z+d=42/6

or z+d=7

or 10y=7

or y=7/10

Then,

7/10=z+d

I cannot finish it but

using the above given ideas you can find the solution easily.

Hope this will help you.