Question

The digits of a three-digit positive integer are in A.P. and their sum is $15$. The number obtained by reversing the digits is $594$ less than the original number. Then the unit place of the number is

• A. $4$
• B. $2$
• C. $7$
• D. $5$

Answer 1

The correct option is B $2$

Let the digits at ones, tens and hundreds place be $(a-d),a$ and $(a+d)$ respectively.

Then the number
$=(a+d)\times 100+(a\times 10)+(a-d)\times 1=111a+99d$

The number obtained by reversing the digits
$=(a-d)\times 100+(a\times 10)+\left(\right(a+d)\times 1)=111a-99d$

It is given that
$(a-d)+a+(a+d)=15$
and $111a-99d=111a+99d-594$
$\Rightarrow 3a=15\Rightarrow a=5$
and $198d=594\Rightarrow d=3$

Hence, the required number is $111\times 5+99\times 3=852$, so the unit place number is $2$.