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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
4 years, 3 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.

- 4 years, 3 months ago

Thanks again.

- 4 years, 3 months ago

You're welcome.

- 4 years, 3 months ago

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.)

- 4 years, 3 months ago