Question

A number consists of three digits which are in GP the sum of the right-hand digits exceeds twice the middle digits by $1$ and the sum of the left-hand and middle digits is two-third of the sum of the middle and right-hand digits. Find the number.

Answer 1

Solution.

Let the three digits be $a,ar$ and $a{r}^2$.

Then, according to the question,
$\begin{array}{l}100a+10ar+a{r}^2\dots \left(\mathrm{i}\right)\\ a+a{r}^2=2ar+1\\ a\left(r^2-2r+1\right)=1\\ a(r-1{)}^2=1\dots (\mathrm{ii})\\ a+ar=\frac{2}{3}\left(ar+a{r}^2\right)\\ \Rightarrow 3+3r=2r+2r^2\\ \Rightarrow 2r^2-r-3=0\\ \Rightarrow (r+1)(2r-3)=0\\ \therefore r=-1,\frac{3}{2}\end{array}$

For $r=-1$

$\begin{array}{l}a=\frac{1}{(-1-1{)}^2}\\ =\frac{1}{4}\notin 1r\ne -1\end{array}$.

For $r=\frac{3}{2}$

$\begin{array}{l}a=\frac{1}{{\left({\displaystyle \frac{3}{2}}-1\right)}^2}\\ =\frac{1}{4}\notin 1r\ne -1\end{array}$.

From equation (i) the number is

$400+10\times 4\times \frac{3}{2}+4\times \frac{9}{4}=469$

Hence, the number is $469$.