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Perfect Powers

How many \(3\) digit numbers \(N\) can be expressed as \(\overline{aba} = p^q\), where \(p\) and \(q\) are positive integers and \(q>1\)

\(A\) \(request\) : Please explain your steps a bit if you use Modular Arithmetic because I ain't much used to it.

Note by Abhimanyu Swami
3 years, 11 months ago

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Again I believe there is no better choice than a case-by-case analysis. I rephrase the problem to find the number of 3-digit palindrome perfect power. First note that no power of 2 is a 3-digit palindrome. Since \(3^7>1000\), we just need to check until \(q=6\). Furthermore, if \(q\) is even, \(p^q\) is a perfect square and we may assume \(q=2\). Now we consider \(q=5,3,2\).

Case 1: \(q=5\)

\(4^5>1000\), and \(3^5\)is not a palindrome.

Case 2: \(q=3\)

Just manually check for \(1\leq p\leq 9\), and we find only \(7^3=343\) is a palindrome.

Case 3: \(q=2\)

Recall that the quadratic residues modulo \(10\) are \(1,4,9,6,5\). Substitute \(a=1,4,9,6,5\) and check to find a \(b\) such that \(\overline{aba}\) is a square. One arrives at \(11^2=121, 22^2=484, 26^2=676\).

Therefore we conclude there are 4 3-digit palindrome perfect powers.

Yong See Foo - 3 years, 11 months ago

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Thanks again.

Abhimanyu Swami - 3 years, 11 months ago

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You're welcome.

Yong See Foo - 3 years, 11 months ago

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The answer is 4. (7^3 = 343, 11^2 = 121, 22^2 = 484, 26^2 = 676) *Brute force. (I'll try to prove this later.)

John Ashley Capellan - 3 years, 11 months ago

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