Question

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

Answer 1

Let the digits of the number be (a - d), (a), (a + d) (ones, tens and hundred, respectively),
then the number so formed
= 100 (a + d) + 10 (a) + 1 (a - d)
= 100a + 100d + 10a + a - d
= 111a + 99d . . . (i)
Number formed by reversing the digits
= 100(a - d) + 10 $\times$ a + 1(a+ d)
= 100a - 100d + 10a + a + d
= 111a - 99d . . . . .(ii)
Now, according to the question,
a + d + a + a - d = 21
$\Rightarrow$ 3a = 21 $\Rightarrow$ a = 7
Also,
111a + 99d = 111a - 99d + 396 [from Eqs. (i) and (ii)]
$\Rightarrow$ 198d = 396
$\Rightarrow$ d = 2
$\therefore$ The number is 975. [from Eq. (i)]