Let $N = \overline{a_1 a_2 a_3 \cdots a_n}$ be the number we are trying to find. We can write the condition in the problem as

$4\overline{a_1 a_2 a_3 \cdots a_n} = \overline{a_n a_1 a_2 \cdots a_{n-1}}.$

Let $A = \overline{a_1 a_2 a_3 \cdots a_{n - 1}}$ and $B = a_n$ . We have

$\begin{aligned} 4(10A + B) &= 10^{n - 1} B + A \\ 40A + 4B &= A + 10^{n - 1} B \\ 39A &= B(10^{n - 1} - 4) \\ 39(a_1 \times 10^{n - 2} + \cdots + a_{n - 1}) &= a_n(10^{n - 1} - 4). \end{aligned}$

Now, since $a_n$ and $10^{n - 1} - 4$ are both integers, we must have that $a_n(10^{n - 1} - 4)$ divides $39$ . This is equivalent to saying that $a_n(10^{n - 1} - 4)$ is divisible by both $3$ and $13$ . We know that

$10^{n - 1} - 4 \equiv 1^{n - 1} - 1 \equiv 0 \! \! \! \! \pmod{3},$

so either $13|a_n$ or $13|10^{n - 1} - 4$ . If $13|a_n$ , then $a_n \geq 13$ , which cannot happen since $a_n$ is a digit. Thus, we must have $13|10^{n - 1} - 4$ , or $10^{n - 1} \equiv 4 \pmod{13}$ . Checking powers of $10$ mod $13$ , we find that $10^{5 + 6k} \equiv 4 \pmod{13}$ , for all nonnegative integers $k$ . We want the smallest value of $n$ , so we must have $n - 1 = 5$ , or $n = 6$ .

Now that we know the value of $n$ , we can plug it into our original equation and divide both sides by $39$ :

$\begin{aligned} 39(a_1 \times 10^4 + \cdots + a_5) &= a_6(10^5 - 4) \\ 39(a_1 \times 10^4 \cdots + a_5) &= 99996a_6 \\ a_1 \times 10^4 + a_2 \times 10^3 + a_ 3 + 10^2 + a_4\times 10 + a_5 &= 2564a_6. \end{aligned}$

The left hand side is a five digit number, so we find the smallest value of $a_6$ such that $2564a_6$ is a five digit number, which will give the smallest such five digit number. This yields $a_6 = 4$ , and $\overline{a_1a_2a_3a_4a_5} = 2564 \times 4 = 10256.$ Thus, $N = \overline{a_1a_2a_3a_4a_5a_6} = \boxed{102564}.$

We can check easily that $102564 \times 4 = 410256$ , as desired.

include<iostream.h>

#include<conio.h>

#include<math.h>

unsigned long int newnum (unsigned long int a)

{

unsigned long int lim=0,num2=0,m,digit,count=0;

m=a;

while(a>0)

{

    lim+=1;

    a=a/10;

}

a=m;

while(count<lim)

{

     digit=a%10;

     if(count==0)

        num2+=digit*pow(10,lim-1);

     else

        num2+=digit*pow(10,count-1);

     a=a/10;

     count+=1;

}

a=m;

return num2;

}

void main()

{

clrscr();

unsigned long int a=1,x,y;

while(a<200000)

{

    y=a%10;

    x=newnum(a);

    if(x==4*a&&y!=0)

    {

        cout << a;

        break;

    }

    a=a+1;



}

getch();

}

I don't know why latex is acting up. This is the first time I'm posting a programming solution. And I'm not sure it's fair to do so here. Would really appreciate if a mod could do the needful here. There's a # in the beginning which somehow doesn't show up before people start pointing out.

Program runs in 10 secs to give the answer as $102564$