A positive integer is said to be "nefelibata" if, upon taking its last digit and placing it as the first digit, keeping the order of all the remaining digits intact (for example, 312 -> 231), the resulting number is exactly double the original number. Find the smallest possible nefelibata number.
Problem
Source: OLCOMA (Costa Rican MO) Shortlist, 2017
Tags: number theory, Digits
16.05.2018 22:42
i found $105263157894736842$ \
16.05.2018 22:42
You mean uglier than it looks?
16.05.2018 22:44
16.05.2018 23:18
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 $52631578947368421$.
16.05.2018 23:51
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!
16.05.2018 23:53
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, $105263157894736842$.
16.05.2018 23:55
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 $2$.
17.05.2018 01:40
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.