
4. Find all three digit natural numbers of the form $(a b c)_{10}$ such that $(a b c)_{10},(b c a)_{10}$ and $(c a b)_{10}$ are in geometric progression. (Here $(a b c)_{10}$ is representation in base 10.)


## Solution

Let us write

$$
x=\left(a \times 10^{2}\right)+(b \times 10)+c, \quad y=\left(b \times 10^{2}\right)+(c \times 10)+a, \quad z=\left(c \times 10^{2}\right)+(a \times 10)+b
$$

We are given that $y^{2}=x z$. This means

$$
\left(\left(b \times 10^{2}\right)+(c \times 10)+a\right)^{2}=\left(\left(a \times 10^{2}\right)+(b \times 10)+c\right)\left(\left(c \times 10^{2}\right)+(a \times 10)+b\right)
$$

We can solve for $c$ and get

$$
c=\frac{10 b^{2}-a^{2}}{10 a-b}
$$

If $a, b, c$ are digits leading to a solution, and if $d=\operatorname{gcd}(a, b)$ then $d \mid c$. Consequently, we may assume that $\operatorname{gcd}(a, b)=1$. Now

$$
c=\frac{999 a^{2}}{10 a-b}-(10 b+100 a)
$$

showing that $10 a-b$ divides $999 a^{2}$. Since $a, b$ are relatively prime, this is possible only if $10 a-b$ is a factor of 999 . It follows that $10 a-b$ takes the values $1,3,9,27,37$. These values lead to the pairs

$$
(a, b)=(1,9),(1,7),(1,1),(4,3)
$$

We can discard the first two pairs as they lead to a value of $c>10$. The third gives the trivial solution $(111,111,111)$. Taking $d=2,3,4,5,6,7,8,9$, we get 9 solution:

$$
(a b c)_{10}=111,222,333,444,555,666,777,888,999
$$

The last pair gives $c=2$ and hence the solution $(432,324,243)$. Another solution is obtained on multiplying by 2 : $(864,648,486)$.

Thus we have

$$
(a b c)_{10}=111,222,333,444,555,666,777,888,999,432,864
$$
