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Here is the problem:

Can you determine the smallest natural number $N>0$ not divisible by $10$, such that when you move the last two digits of $N$ to the front, shifting the other digits two places to the right, you end up with $6N$?

For example $N=1234567$ will not work, since $6N=7407402$ whereas moving the last two digits in front results in $6712345$. It is not allowed to fill in zeros in case that helps, so $67012345$ is not an option in the current example. Also note, that the last two digits must remain **in the same order** as they originally were. Also you can only use base 10 numbers.

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The current question was inspired by this other question found on this site.

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If we take $N=100b+a$ with $10^{n-1} < b < 10^n$, then the specified condition is:

$$\begin{align}\\

a10^n+b &=6(100b+a) \\

(10^n-6)a &=599b\\

\end{align}$$

So $599$ is prime and therefore we need $599 \mid (10^n-6)$, that is, $10^n \equiv 6 \bmod 599$. This happens first for $n=297$. To bring $a$ into range, and avoid multiples of $10$, we need to multiple by $61$. In principle, then, we will need $N = \dfrac{6100(10^{297}-6)}{599}+61$

Check the digits out here

I solved the reverse problem first, by mistake. In this version, I shift the first two digits of $N$ to the end, and that gives six times $N$.

$$\begin{align}\\

6(a10^n+b) &=100b+a \\

(6\cdot 10^n-1)a &=94b\\

\end{align}$$

We cannot have $b=(6\cdot 10^n-1)$ so we must have $94 \mid (6\cdot 10^n-1) \implies 47 \mid (6\cdot 10^n-1)$

So the smallest value for which this is feasible is $n=44$, where:

$(6\cdot 10^n-1)/47 = 12765957446808510638297872340425531914893617$

In order for $a$ to have two digits, we need $b$ to be $5$ times this number (giving $a=10$), and

$$N=1063829787234042553191489361702127659574468085$$

First thing to note: your $N$ and $6N$ have the same number of digits.

Since we’re working in base 10, this means we know that our number is less than or equal to $16…6$ and greater than $10…0$, since if we multiple $9…9$ (one less digit than the numbers before) by $6$, we’ll get one digit more than $9…9$ has.

That’s a fairly small window, and also, it gives us the first digit of $N$ as $1$. This narrows down our search for the first digit of $6N$ also: it can be no less than $6$. Thus, the 5th digit of $N$ can be no less than $6$.

Now, let’s say $N = 100b + a$ for $60 < a < 100$ and $1000 < b < 1666$.

Then we also know $6N = 10000a + b$. Then we have that $9999a – 99b$ is divisble by 5, so $101a – b$ is divisible by $5$, thus $a – b$ is divisible by $5$. This tells us that the 6th digit of $N$ is either exactly the 4th digit of $N$ or their difference is $5$.

Let $N$ have $m+1$ digits. We have $N= 100x+10a+b$, where $x$ has $m-1$ digits. Moving the last two digits in front we obtain,

$$6N = 10^ma + 10^{m-1}b + x \implies 600x + 60a + 6b = 10^ma + 10^{m-1}b + x$$

Hence, we need

$$599x = (10^m-60)a + (10^{m-1}-6)b = (10^{m-1}-6)(10a+b)$$

This means $599$ divides $(10^{m-1}-6)(10a+b)$. Further, $10a+b$ is a two digit number and $599$ is a prime. Hence, we need $m$ such that $599$ divides $10^{m-1}-6$. This gives us $m=298,597,\ldots$. I trust you can finish it off from here.

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