Question

The digits of a three digit natural number are in AP and their sum is The number obtained by reversing the digits is 396 less than the original number. Find the number.​

Answer 1

Let the numbers be,

a−d,a,a+d

Given,

a−d+a+a+d=15

3a=15

∴a=5

(a−d)100+10a+(a+d)=(a+d)100+10a+(a−d)+396

100a−100d+10a+a+d=100a+100d+10a+a−d+396

−198d=396

∴d=−2