Prove that there exists a power of three that ends with 001

Solution


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Copyright © 1996-2018 Alexander Bogomolny

Prove that there exists a power of three that ends with 001

As in a related problem, let 3n and 3m (n > m) have the same remainder when divided by 1000. Thus 3n - 3m = 3m(3n-m - 1) is divisible by 1000. Since 1000 and 3m have no common factors, 1000 is bound to divide the second factor (3n-m - 1). This exactly means that 3n-m ends with 001!


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    Copyright © 1996-2018 Alexander Bogomolny

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