HMMT February

Number theory Difficulty 5.1 AIME, harder Prove it United States

Problem:

Mark writes the expression $\sqrt{abcd}$ on the board, where $abcd$ is a four-digit number and $a \neq 0$ . Derek, a toddler, decides to move the $a$ , changing Mark's expression to $a\sqrt{bc}$ . Surprisingly, these two expressions are equal. Compute the only possible four-digit number $abcd$ .

Solution

Solution:

Let $x = bcd$ . Then, we rewrite the given condition $\sqrt{abcd} = a\sqrt{bc}$ as

$1000a + x = a^{2}x,$

which simplifies as

$(a^{2} - 1)x = 1000a.$

In particular, $a^{2} - 1$ divides $1000a$ . Since $\gcd(a^{2} - 1, a) = 1$ , it follows that $a^{2} - 1 \mid 1000$ . The only $a \in \{1, 2, \ldots, 9\}$ that satisfies this is $a = 3$ . Then $8x = 3000$ , so $x = 375$ . Thus $abcd = \boxed{3375}$ .

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