8 Shift

Just approach this problem like long multiplication.

$\large{\begin{array}{ccccccc} && & & & & \square & 8\\ \times && & & & & &2\\ \hline & & & & & &&\square &\\ \end{array}}$ The bottom box is a 6 which we can then write in the upper box. Continue this process (don't forget to carry the 1's) until you reach an 8 (without a carrying 1). This will give you the smallest possible solution of 421052631578947368.

If the original number is $10x+8,$ then twice that number is $8 \cdot 10^n + x,$ where $n$ is the number of digits of $x.$ This leads to the equation $8(10^n-2) = 19x.$ The smallest solution will use the smallest positive $n$ such that $10^n \equiv 2$ mod $19.$ Looking at powers of $10$ mod $19,$ we see that $10$ is a primitive root , and the smallest $n$ such that $10^n \equiv 2$ mod $19$ is $n=17.$ Checking $x = \frac{8(10^{17}-2)}{19},$ we see that it does indeed have $17$ digits, so the answer is $10x+8 = \fbox{421052631578947368}.$

To phrase the problem mathematically, $\frac{x-8}{10} + 8 \times 10^{n} = 2x$ , where x is our original number, and n is the amount of digits in our original number. $x-8+8 \times 10^{n+1}=20x$

$8(10^{n+1}-1)=19x$

$10^{n+1}-1=\frac{19}{8}x$ (1)

The left-hand side of this equation evaluates to $9 \times$ a repunit of 1. As x is an integer, and clearly, 19 does not divide 9, we can see that 19 must divide the repunit of 1. A quick calculator trick can yield the smallest repunit that has this property.

$19 \times 5847953216374269=111111111111111111$ , a repunit with 18 digits

This tells us that $10^{n+1}-1=111111111111111111 \times 9$

$10^{n+1}=10^{19}$

$n=18$

Sub into (1)

$(10^{19}-1)\frac{8}{19}=x$

$x=421052631578947368$

And indeed, $\frac{842105263157894736}{2}=421052631578947368$

I would be interested to hear other solutions to this problem, preferably ones that do not require calculator tomfoolery

Let $n$ be the desired integer. We will use $\Vert$ to represent concatenation. We have $n=x\Vert8$ and $2n=8\Vert{x}$ for some integer $x$ . We can rewrite these expressions as $n=10x+8$ and $2n=8*10^a+x$ for some integer $a$ . We perform some basic algebra: $2n=8*10^a+x\\ 2(10x+8)=8*10^a+x\\ 20x+16=8*10^a+x\\ 19x=8*10^a-16\\ 19x=8*(10^a-2)\\ \frac{19}{8}x=10^a-2.$ Since $10^a-2$ is an integer, $\frac{19}{8}x$ is also an integer. Since $8$ and $19$ are relatively prime, $b=\frac{x}{8}$ is an integer. So we have $19b=10^a-2$ , or equivalently, $10^a \equiv 2 \pmod{19}$ . Multiplying by $10$ , we get $10^{a+1} \equiv 1 \pmod{19}$ . Since $19$ is prime, we can use Fermat's Little Theorem to find the smallest such $a$ that satisfies this congruence. Then, $a+1=18$ , so $a=17$ . (Presumably, we could find some extremely large integers for which the desired shifting property holds by instead setting $a=18^2-1,18^3-1,...$ .)

From here, it is a simple matter of solving for $n$ : $n=10x+8\\ n=10(8b)+8\\ n=10(8 * \frac{10^a-2}{19})+8\\ n=10(8 * \frac{10^{17}-2}{19})+8\\ n=\boxed{421052631578947368}.$

The following code solves the problem more generally for the question $\overline{abc\dots X} \times F = \overline{Xabc\dots},$ for any factor $F$ and any digit $X$ . The code stops after too many digits have been generated; it does not actually check if a solution is reached, but that can easily be mended.

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#include<stdio.h>

#define MAX_DIGIT 30
#define LAST_DIGIT 8
#define FACTOR 2

int main() {
    int d[MAX_DIGIT], c[MAX_DIGIT];
    int n = MAX_DIGIT;

    --n; d[n] = LAST_DIGIT; c[n] = 0;
    do {
        int nd = FACTOR*d[n] + c[n];
        --n; d[n] = nd % 10; c[n] = nd / 10;
        if (d[n] == LAST_DIGIT && c[n] == 0) break;
    } while (n > 0);

    n++;
    while (n < MAX_DIGIT) printf("%d", d[n++]);
    printf("\n");
}

Output:

1

421052631578947368

A general approach can be as follows: $$ Let the $n$ digit number be $10r+d$ ---------------------- $\textbf{(1)}$ $$ After the last digit is moved to the first , the number looks like $10^{n+1}d+r$ ---------------------- $\textbf{(2)}$ . $$ By Question, $2\times(10r+d) = 10^{n+1}d+r$ Solving for $d=8$ in the abobe expression we get $r=\frac{8.10^{n+1}-16}{19}$ Since $19$ is prime so we have the following set of congruences: $8.10^{n+1} \equiv 16 \mod 19$ $10^{n+1} \equiv 2 \mod 19$ $5.10^n \equiv 1\mod19$ $10^n \equiv 4\mod19$ $25.10^{n-2}\equiv 1\mod19$ $10^{n-2} \equiv 16 \mod19$ $25.10^{n-4} \equiv 4\mod19$ $625.10^{n-6} \equiv 1\mod19$ $10^{n-6} \equiv 9\mod19$ $10^{n-7} \equiv 18 \mod19$

Now starting with $p=2$ and repeatedly applying the congruences modulo 19 i.e, $10^p \mod 19$ with increasing $p$ . $$ We see that $p=9$ satisfies the last congruence in the above set of congruences(you could have satisfied any of the congruences in the above set but I chose the last one as it required the least effort) So, substituting $9=p=n-7$ we get $n=16$ which is indeed the solution. Substitute this $n$ in $2$ to get $r$ and in $1$ substitute $r$ to get the original number

This is a "Mind Your Decisions" video that shows both the long multiplication solution John Ross showed and the mod 19 solution Patrick Corn showed (by letting a in the video to equal 8).