# Does there exist a positive integer $n$ such that it will be twice of $n$ when its digits are reversed?

Does there exist a positive integer $n$ such that it will be twice of $n$ when its digits are reversed?

We define $f(n)=m$ where the digits of $m$ and $n$ are reverse.
Such as $f(12)=21,f(122)=221,f(10)=01=1$,so we cannot say $f(f(n))=n$,but $f(f(n))=n/10^k$.

So we need to find a solution to $f(n)=2n$.

If $f(n)=2n$ and the first digit of $n$ is 2,then the last digit of $n$ is 1 or 6,and so on.So the first digit of $n$ is even.

There are some solutions to the equation $f(n)=\frac{3}{2}n$,such as $n=4356,43956$,but there is no solution to $f(n)=2n$ when $n<10^7$.

Edit:Since Alon Amit has proved that $f(n)=2n$ has no positive solution,so I wonder whether $f(n)=\frac{3}{2}n$ has only finite solutions.

Any suggestion is grateful,thanks in advance!

#### Solutions Collecting From Web of "Does there exist a positive integer $n$ such that it will be twice of $n$ when its digits are reversed?"

There is no such integer $n$.

Suppose there is, and let $b = n \bmod 10$ be its units digit (in decimal notation) and $a$ its leading digit, so $a 10^k \leq n < (a+1)10^k$ for some $k$ and $1 \leq a < 10$.

Since $f(n) = 2n$ is larger than $n$, and $f(n)$ has leading digit $b$ and at most as many digits as $n$, we must have $b > a$. At the same time, $2b \equiv a \bmod 10$ because $(2b \bmod 10)$ is the units digits of $2n$ and $a$ is the units digit of $f(n)$.

This means that $a$ is even, as you pointed out.

• $a$ cannot be $0$, by definition.
• If $a=2$, $b$ must be $1$ (impossible since $b>a$) or $6$. But the leading digit of $2n$ can only be $4$ or $5$, since $4\times 10^k \leq 2n < 6\times 10^k$ and the right inequality is strict (in plain English, when you double a number starting with $2$, the result must start with $4$ or $5$).
• If $a=4$, by the same reasoning $b$ must be $7$ which again fails to be a possible first digit for twice a number whose leading digit is $4$.
• If $a=6$, we have $b=8$. Impossible since $2n$ must start with $1$.
• If $a=8$, $b$ must be $9$. Impossible again, for the same reason.

So no $a$ is possible, QED.

Edit: The OP has further asked if $f(n) = \frac{3}{2}n$ has only finitely many solutions. The answer to that is No: Consider $n=43999…99956$ where the number of $9$’s is arbitrary. One can check that $f(n) = \frac{3}{2}n$ for those values of $n$.