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.$\left[3Marks\right]$

Answer 1

Let the digits of the number be (a - d), (a), (a + d) (ones, tens and hundred, respectively),$\left[0.5Mark\right]$

Then the number so formed
= 100 (a + d) + 10 (a) + 1 (a - d)
= 100a + 100d + 10a + a - d
= 111a + 99d . . . (i)$\left[0.5Mark\right]$

Number formed by reversing the digits
= 100(a - d) + 10 $\times$ a + 1(a+ d)
= 100a - 100d + 10a + a + d
= 111a - 99d . . . . .(ii)$\left[0.5Mark\right]$

Now, according to the question,
a + d + a + a - d = 21
$\Rightarrow$ 3a = 21 $\Rightarrow$ a = 7$\left[0.5Mark\right]$

Also,
111a + 99d = 111a - 99d + 396 [from Eqs. (i) and (ii)]
$\Rightarrow$ 198d = 396
$\Rightarrow$ d = 2$\left[0.5Mark\right]$

$\therefore$ The number is 975. [from Eq. (i)]$\left[0.5Mark\right]$