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Fermat numbers (Posted on 2017-07-13) Difficulty: 1 of 5
For any non-negative integer n define f(n)=2^(2^n)+1.
The above formula produces Fermat numbers i.e. 3,5,17,257 ...etc

Prove: Except the 1st term, the digital root of sequence's members are either 5 or 8, depending on parity of n.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (2 votes)

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Solution Answer Comment 2 of 2 |
2^2=4
4^2=16=7 mod 9
7^2=49=4 mod 9
Therefore, 2^(2^n) mod 9 is 2, 4, 7, 4, 7, ... Then, 2^(2^n)+1 mod 9 is 3, 5, 8, 5, 8, ...


  Posted by Math Man on 2017-07-19 14:22:52
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