i found $105263157894736842$ \

You mean uglier than it looks?

hint

Write the original number as $10A+B$, where $A$ has $n$ digits. So we need to find positive integers $a$,$b$,$n$, such that $10^nB+A=20A+2B$, and $B$ is less than $10$. Rearranging, we get $19A=\left(10^n-2\right)B$. As $B$ cannot have a factor of $19$, we need to find the smallest $n$ such that $10^n-2$ is divisible by $19$. This $n$ is $17$. So we have $19A=\left(10^{17}-2\right)B$. To minimise the original number, we want $A$ to be as small as possible, so we want $B$ to be as small as possible. Setting $B=1$, we get the smallest possible such number as $10\cdot\frac{10^{17}-2}{19}+1$, or $52631578947368421$.

Hamel wrote: You mean uglier than it looks? Yes lol; we were all shocked when we saw the ugliness of the computations necessary. That's why it wasn't put on the exam. However, I do think it's a cool problem, if only because I find it incredibly interesting that the first one that works is that huge, ythomashu wrote: i found $105263157894736842$ \ Nice!

Kaskade wrote: Write the original number as $10A+B$, where $A$ has $n$ digits. So we need to find positive integers $a$,$b$,$n$, such that $10^nB+A=20A+2B$, and $B$ is less than $10$. Rearranging, we get $19A=\left(10^n-2\right)B$. As $B$ cannot have a factor of $19$, we need to find the smallest $n$ such that $10^n-2$ is divisible by $19$. This $n$ is $17$. So we have $19A=\left(10^{17}-2\right)B$. To minimise the original number, we want $A$ to be as small as possible, so we want $B$ to be as small as possible. Setting $B=1$, we get the smallest possible such number as $10\cdot\frac{10^{17}-2}{19}+1$, or $52631578947368421$. Actually, this is wrong. Setting B=1 doesn't work, as the resulting number has only 16 digits, which cannot work, as we determined n=17. The answer above is right: setting B=2 gives the desired result, $105263157894736842$.

GeometryIsMyWeakness wrote: Kaskade wrote: Write the original number as $10A+B$, where $A$ has $n$ digits. So we need to find positive integers $a$,$b$,$n$, such that $10^nB+A=20A+2B$, and $B$ is less than $10$. Rearranging, we get $19A=\left(10^n-2\right)B$. As $B$ cannot have a factor of $19$, we need to find the smallest $n$ such that $10^n-2$ is divisible by $19$. This $n$ is $17$. So we have $19A=\left(10^{17}-2\right)B$. To minimise the original number, we want $A$ to be as small as possible, so we want $B$ to be as small as possible. Setting $B=1$, we get the smallest possible such number as $10\cdot\frac{10^{17}-2}{19}+1$, or $52631578947368421$. Actually, this is wrong. Setting B=1 doesn't work, as the resulting number has only 16 digits, which cannot work, as we determined n=17. The answer above is right: setting B=2 gives the desired result, $105263157894736842$. It is the same answer, just multiply my result by $2$.

I would just bash it out like so: Take a starting digit (say 1). Then multiply by 2. We get 2 so the next digit is a 2. Now we have 21 Multiply by 2, the next digit is a 4. 421 Repeat... 8421 68421 368421 7368421 46368421 946368421 8946368421 78946368421 578946368421 1578946368421 This starts with a $1$, so the answer is the previous number, $\boxed{578946368421}$ Unfortunately, we still have to check if you start with a different digit. EDIT: Wait oops, you still have to continue as you have carried a digit. I think you actually have to start with a different digit for it to ever end.