Perfect Powers

How many 33 digit numbers NN can be expressed as aba=pq\overline{aba} = p^q, where pp and qq are positive integers and q>1q>1

AA requestrequest : Please explain your steps a bit if you use Modular Arithmetic because I ain't much used to it.

Note by Abhimanyu Swami
6 years, 10 months ago

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4 votes

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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 37>10003^7>1000, we just need to check until q=6q=6. Furthermore, if qq is even, pqp^q is a perfect square and we may assume q=2q=2. Now we consider q=5,3,2q=5,3,2.

Case 1: q=5q=5

45>10004^5>1000, and 353^5is not a palindrome.

Case 2: q=3q=3

Just manually check for 1p91\leq p\leq 9, and we find only 73=3437^3=343 is a palindrome.

Case 3: q=2q=2

Recall that the quadratic residues modulo 1010 are 1,4,9,6,51,4,9,6,5. Substitute a=1,4,9,6,5a=1,4,9,6,5 and check to find a bb such that aba\overline{aba} is a square. One arrives at 112=121,222=484,262=67611^2=121, 22^2=484, 26^2=676.

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

Yong See Foo - 6 years, 10 months ago

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Thanks again.

Abhimanyu Swami - 6 years, 10 months ago

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You're welcome.

Yong See Foo - 6 years, 10 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 - 6 years, 10 months ago

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