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Harvard-MIT Mathematics Tournament

Number theory Difficulty 5.4 AIME, harder Prove it United States

Problem:
Call a positive integer in base $10$ $k$ -good if we can split it into two integers $y$ and $z$ , such that $y$ is all digits on the left and $z$ is all digits on the right, and such that $y = k \cdot z$ . For example, $2010$ is $2$ -good because we can split it into $20$ and $10$ and $20 = 2 \cdot 10$ . $20010$ is also $2$ -good, because we can split it into $20$ and $010$ . In addition, it is $20$ -good, because we can split it into $200$ and $10$ .
Show that there exists a $48$ -good perfect square.

Solution

Solution:
We wish to find integers $a$ , $z$ such that $48z \cdot 10^{a} + z = z\left(48 \cdot 10^{a} + 1\right)$ is a perfect square, where $z < 10^{a}$ . This would prove that there exists a $48$ -good perfect square because we are pulling off the last $a$ digits of the number and get two integers $48z$ and $z$ .

To make $z$ small by keeping the product a perfect square, we'd like $48 \cdot 10^{a} + 1$ to be divisible by some reasonably large square. Take $a = 42 = \varphi(49)$ . By Euler's theorem, $10^{42} \equiv 1 \pmod{49}$ , so $48 \cdot 10^{a} + 1$ is a multiple of $49$ .

Then we can take $z = \frac{48 \cdot 10^{a} + 1}{49}$ . (Clearly $z < 10^{a}$ , so we're fine.) Then we have
$z\left(48 \cdot 10^{a} + 1\right) = \left(\frac{48 \cdot 10^{42} + 1}{7}\right)^{2}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.