Double adding digits

No. $(A, B) = {(3, 6), (13, 52)}$ are the only solutions.

Proof: Assume that the number of digits in $B$ is $n$ . Then $\overline{AB} = 10^n \cdot A + B = 2\cdot AB \Leftrightarrow B(2A-1) = 10^n \cdot A$

And since $gcd(A, A-1) = gcd(A, A-1 +A) = 1$ therefore $2A-1 \mid 10^n$ but $2A-1$ is odd so it must be a divisor of $5^n$ which implies that $2A-1 \leq 5^n \Leftrightarrow A \leq \frac{5^n - 1}{2}$

On the other hand we know that the number of digits in both $A, B$ is equal to $n$ therefore $A$ cannot be less than $10^{n-1}$ and from the previous inequality we get: $10^{n-1} \leq A \leq \frac{5^n - 1}{2}$

$\Leftrightarrow 5^n - 2^n \cdot 5^{n-1} \geq 1 \Leftrightarrow 5^{n-1} \cdot (5-2^n) \geq 1$ and from here we find that $n$ cannot be big because the $LHS$ must be positive, the only possible values for $n$ here are only 1 and 2. From which we get the solutions mentioned above for $A, B$ .

Let $n$ be the number of digits of the numbers. Then we want to solve the diophantine equation $2ab = 10^na + b$

Lemma: $b \equiv 0 \mod 2^n$

From the equation it is clear that $b \equiv 0 \mod 2$ . Now suppose $b \equiv 0 \mod 2^k$ , thus $2ab \equiv 0 \mod 2^{k+1}$ and if $k < n$ , $b \equiv 0 \mod 2^{k+1}$ . By induction, $b \equiv 0 \mod 2^n$ .

Thus, we may write $b = 2^n b_0$ and write the equation as $2^{n+1} a b_0 = 10^n a + 2^n b_0 \implies 2ab_0 = 5^n a + b_0 \implies b_0 \equiv 0 \mod a$ . so we may write $b_0 = ab_1$ . From this we can write $b = 2^n a b_1$ . With this the equation becomes $2a^2 b_1 = 5^n a + a b_1 \implies 2ab_1 = 5^n + b_1 \implies a = \frac{5^n + b_1}{2b_1}$ and $b = 2^{n-1} (5^n + b_1)$ .

To solve the problem, we will find a property of valid $b_1$ . From the equation $2ab_0 = 5^na + b_0$ we deduce $5^n a \equiv 0 \mod b_0 \implies 5^n a \equiv 0 \mod ab_1 \implies 5^n \equiv 0 \mod b_1$ . This means that either $b_1 = 1$ or $5 | b_1$ . Considering the first case, $a = \frac{5^n + 1}{2}$ . For $n=1,2$ we find the solutions given in the problem. However, if we try $n = 3$ we find that $a$ has 2 digits, not 3. And indeed, it turns out that $\frac{5^n + 1}{2} < 10^{n-1}$ for $n \geq 3$ . Thus, for $b_1 = 1$ there are no other solutions than the ones given in the problem.

Now, if we assume that $5 | b_1$ we find another contradiction. Suppose that $a = \frac{5^n + b_1}{2b_1} \geq 10^{n-1}$ . Then $b = 2^{n-1} (5^n + b_1) \geq 2^{n-1} \times 10^{n-1} 2b_1 \geq 2^n 10^{n-1} \times 5 \geq 10^n$ . So, if $a$ has $n$ digits, $b$ has at least $n+1$ digits. This contradicts that they must have the same amount of digits.