Numbers in Different Bases (reversed representation after conversions) The problem is to find a 3-digit natural number in base ten, where its depiction in base seven is the reverse of its depiction in base nine. 248 is one answer, but then there remains the fact that a 3-digit natural number in base ten can become a 4-digit natural number in base seven or nine. It would be really appreciated if someone could explain how 248 is the only possible answer and that it doesn't work when the 3-digit natural number in base ten becomes a four digit natural number in base seven or nine.
Thanks :)
 A: For the case of a 3 digits number in bases $7$ and $9$, the answer is easy : let $a$, $b$ and $c$ be the digits of those numbers. Then you have to solve the diophantine equation 
$$a\times9^2+b\times9+c=c\times7^2+b\times7+a\iff 40a+b-24c=0$$
This means 
$$b=8(3c-5a)$$
so $8$ must divide $b$. But $0\le b\le 6$ (it is a digit of a base 7 number), hence $b=0$. Remains $3c=5a$, and the only possible solution is $c=5$, $a=3$, which leads to the solution 
$$5\times 7^2+3=3\times9^2+5=248$$
For the 4 digits case, the equation is
$$728a+74b-40c-342d=0\iff 364a+37b-20c-171d=0$$
Isolating the biggest coefficient, you find
$$171d-222\le 364a\le 171d+120$$
There are now three possibilities for $a$ :
$$a=1\Longrightarrow d=2\text{ or }3,\quad a=2\Longrightarrow d=4\text{ or }5,\quad a=3\Longrightarrow d=6$$
Just test those possibilities to eliminate solutions. For example : $a=1$, $d=2$ leads to
$$37b+22-20c=0$$
Modulo $5$, this is $2b+2\equiv0\mod5$ so $b\equiv4\mod5$. $b$ must then be $0$ or $4$, none of these cases work. 
A: Suppose the number is
$n = 729a+81b+9c+d = 343d+49c+7b+a$ with $1 \leq a \leq 6,$
$0 \leq b \leq 6,$ $0 \leq c \leq 6,$ and $1 \leq d \leq 6,$
that is, a four-digit number in both bases nine and seven.
Since this must be a three-digit number in base ten,
$729a < n < 1000,$ which implies that $a < 2.$ Therefore $a = 1$
and $n > 729.$
Similarly, $343d < n < 1000$ implies $d < 3.$
But $49c + 7b+a < 343,$ hence $343d + 343 > n > 729$ implies $d > 1.$
Therefore $d = 2.$
That is, $n = 729 + 81b + 9c + 2 = 686 + 49c + 7b + 1,$
from which $40c = 44 + 74b,$
that is, $20c = 22 + 37b.$
First proof
From $c \leq 6$ it follows that $37b = 20c - 22 \leq 98$
and therefore $b \leq 2.$
But then $20c = 22 + 37b \leq 96$ implies $c \leq 4,$
which implies $37b = 20c - 22 \leq 58,$ so $b \leq 1.$
But then $20c = 22 + 37b \leq 59$ implies $c \leq 2,$
which implies $37b = 20c - 22 \leq 18,$ so $b = 0.$
But then $20c = 20, $ which has no integer solution.
Therefore no such number $n$ exists.
Second proof
By trial and error or any other means, find that $20(27) = 22 + 37(14).$
Therefore the solutions of the Diophantine equation
$20c = 22 + 37b$ are $b = 20k+14,$ $c = 37k+27$ where $k$ is an integer.
No such solution satisfies $0 \leq b \leq 6,$ so no such number $n$ exists.
Third proof
From $20c = 22 + 37b$ we conclude $b$ is even.
Let $b = 2m,$ then $20c = 22 + 74m,$ so $10c = 11 + 37m,$
and $0 \equiv 1 - 3m \pmod{10}.$
This has a unique solution, $m \equiv 7 \pmod{10}.$
But $0 \leq b \leq 6$ implies $0 \leq m \leq 3,$ a contradiction.
Therefore no such number $n$ exists.
