perplexus.info :: Sequences : Repunit Crossed Fifth Power Decision

Repunit Crossed Fifth Power Decision (Posted on 2023-03-23)

Consider this infinite sequence which is constituted entirely by repunits:

      11, 111, 1111, 11111, ..........

Does this sequence consist of at least one perfect fifth power?

• If so, provide an example.

• If not, prove that no member of the above mentioned sequence can be a perfect fifth power.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Solution

Comment 1 of 1

The only digit whose 5th power ends in 1 is 1. So the fifth root of the repunit must end in 1.

It is therefore of the form (10a+1)

The last two terms of (10a+1)^5 are 50a+1.

This means and fifth power that ends in the digit 1 also has for its last two digits either 01 or 51.

So no repunit can be a fifth power.

Posted by Jer on 2023-03-23 14:08:36

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