My feeble calculator can't check this

let's define $x$ as $x = \overline {A_2 A_3 \ldots A_{16} }$ , then $10^{14} \le x < 10^{15}$ . Now the equation can be rewritten as: $\left( {10^{15} A_1 + x} \right) \cdot 1.5 = 10x + A_1$ $x = \frac{{1.5 \cdot 10^{15} - 1}}{{10 - 1.5}}A_1$ $x = {\rm 176470588235294} \times A_1$ $x \approx {\rm 1}{\rm .76} \cdot {\rm 10}^{{\rm 14}} A_1$ We know that $A_1$ is a single digit positive integer. Therefore, as long as $10^{14} \le x < 10^{15}$ , any digit is valid. The former condition only holds for the digits $1$ to $5$ , as for $A_1 = 6$ , $x$ becomes greater than $10^{15}$ . In conclusion, there are only $5$ solutions to the equation, and so the answer is $\boxed{5}$ .

The 5 numbers are:

1176470588235294

2352941176470588

3529411764705882

4705882352941176

5882352941176470

Interesting?

Below is the Python 3 program to find the solution to this puzzle.

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def eval(digits):
    value = 0
    for i in range(len(digits)):
        value = 10 * value + digits[i]
    return value


def find_numbers():
    count = 0

    for pivot in range(1, 10):
        carry1 = carry2 = 0
        digits1 = [0] * 16
        digits2 = [0] * 16
        digits1[0] = digits2[-1] = pivot

        for i in range(15, -1, -1):
            target = digits2[i] * 2 + carry1;
            carry1 = target // 10
            target %= 10

            for digit in range(10):
                candidate = 3 * digit + carry2
                if candidate % 10 == target:
                    digits1[i] = digits2[i - 1] = digit
                    carry2 = candidate // 10
                    break

        if 3 * eval(digits1) == 2 * eval(digits2):
            count += 1
            print("Solution with A1 =", pivot, ":", digits1)

    return count


if __name__ == "__main__":
    print("Total numbers found:", find_numbers())