Maths Olympiad Prep

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Number theory Difficulty 5.1 AIME, harder Prove it United States

Problem:

Mark writes the expression abcd\sqrt{abcd} on the board, where abcdabcd is a four-digit number and a0a \neq 0. Derek, a toddler, decides to move the aa, changing Mark's expression to abca\sqrt{bc}. Surprisingly, these two expressions are equal. Compute the only possible four-digit number abcdabcd.

Solution

Solution:

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

1000a+x=a2x,1000a + x = a^{2}x,

which simplifies as

(a21)x=1000a.(a^{2} - 1)x = 1000a.

In particular, a21a^{2} - 1 divides 1000a1000a. Since gcd(a21,a)=1\gcd(a^{2} - 1, a) = 1, it follows that a211000a^{2} - 1 \mid 1000. The only a{1,2,,9}a \in \{1, 2, \ldots, 9\} that satisfies this is a=3a = 3. Then 8x=30008x = 3000, so x=375x = 375. Thus abcd=3375abcd = \boxed{3375}.

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