Question

Let the digits of a three digit number are in G.P. If $400$ is subracted from the number and the digits of the new number are in A.P., then the last digit of the original number is

• A. $8$
• B. $9$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Let the three digits number be $ABC$, where $A=a,B=ar,C=a{r}^2$ and the number be $x$, so
$x=100a+10ar+10a{r}^2$
Now,
$y=x-400\Rightarrow y=100(a-4)+10ar+a{r}^2\cdots \left(1\right)$
Clearly,
$1\le a\le 9\text{and}1\le a-4\le 9\Rightarrow a\in [5,9]\cdots \left(2\right)$

Now, the digits of $y$ are in A.P., so
$2ar=a-4+a{r}^2\Rightarrow a{r}^2-2ar+(a-4)=0\Rightarrow r=\frac{2a\pm \sqrt{4a^2-4a(a-4)}}{2a}\Rightarrow r=\frac{a\pm 2\sqrt{a}}{a}\cdots \left(3\right)$

As $a,ar,a{r}^2$ are integers, so $r$ should be a rational number, therefore $a$ should be a perfect square,
$a=9$
Using equation $\left(3\right)$, we get
$r=\frac{9\pm 6}{9}=\frac{5}{3},\frac{1}{3}$

When $r=\frac{5}{3}$, we get
$ar=15$, which is not possible as $ar\in [1,9]$
When $r=\frac{1}{3}$, we get
$a=9,ar=3,a{r}^2=1$

Hence, the number is $931$.