Problem 1437

1. Let $M$ be a $6n$-digit number that is divisible by $7$. We can express $M$ in the form:
$$M = 10x + y$$
where $x$ is an integer such that $10^{6n-1} \leq x < 10^{6n}$ and $y$ is the last digit of $M$ with $0 \leq y < 10$.

2. When the last digit $y$ of $M$ is moved to the beginning, the new number $N$ can be written as:
$$N = y \cdot 10^{6n-1} + x$$

3. Since $M$ is divisible by $7$, we have:
$$7 \mid M \implies 7 \mid (10x + y)$$

4. By Fermat's Little Theorem, we know that:
$$10^6 \equiv 1 \pmod{7}$$
This implies:
$$10^{6n} \equiv 1 \pmod{7}$$

5. Using the above result, we can rewrite $N$ modulo $7$:
$$N = y \cdot 10^{6n-1} + x$$

6. Since $10^{6n} \equiv 1 \pmod{7}$, we have:
$$10^{6n-1} \equiv 10^{-1} \pmod{7}$$
To find $10^{-1} \pmod{7}$, we need the multiplicative inverse of $10$ modulo $7$. Since $10 \equiv 3 \pmod{7}$, we need the inverse of $3$ modulo $7$. The inverse of $3$ modulo $7$ is $5$ because:
$$3 \cdot 5 \equiv 15 \equiv 1 \pmod{7}$$
Therefore:
$$10^{6n-1} \equiv 5 \pmod{7}$$

7. Substituting this back into the expression for $N$:
$$N \equiv y \cdot 5 + x \pmod{7}$$

8. Since $7 \mid (10x + y)$, we have:
$$10x + y \equiv 0 \pmod{7}$$
This implies:
$$10x \equiv -y \pmod{7}$$
Using $10 \equiv 3 \pmod{7}$, we get:
$$3x \equiv -y \pmod{7}$$
Multiplying both sides by $5$ (the inverse of $3$ modulo $7$):
$$x \equiv -5y \pmod{7}$$

9. Substituting $x \equiv -5y \pmod{7}$ into the expression for $N$:
$$N \equiv y \cdot 5 + (-5y) \equiv 5y - 5y \equiv 0 \pmod{7}$$

10. Therefore, $N$ is divisible by $7$:
$$7 \mid N$$

$\blacksquare$