- #1
Kamataat
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Prove by induction that [tex]2^{2^n}+1[/tex] always ends in 7 for all n > 1 (true for n = 2).
I couldn't figure out anything to do with the last digit being 7, so I looked the case that [tex]2^{2^n}[/tex] ends in 6 for all n > 1, which is also true for n = 2.
Suppose it's true for n = k:
[tex]2^{2^{k+1}}=2^{2^k\cdot 2}=(2^{2^k})^2[/tex]
Now, since by assumption [tex]2^{2^k}[/tex] ends in 6, its 2nd power also ends in 6 since 6x6=36. Hence [tex]2^{2^n}+1[/tex] ends in 7 for all n > 1.
Correct or not?
Thanks in advance,
- Kamataat
I couldn't figure out anything to do with the last digit being 7, so I looked the case that [tex]2^{2^n}[/tex] ends in 6 for all n > 1, which is also true for n = 2.
Suppose it's true for n = k:
[tex]2^{2^{k+1}}=2^{2^k\cdot 2}=(2^{2^k})^2[/tex]
Now, since by assumption [tex]2^{2^k}[/tex] ends in 6, its 2nd power also ends in 6 since 6x6=36. Hence [tex]2^{2^n}+1[/tex] ends in 7 for all n > 1.
Correct or not?
Thanks in advance,
- Kamataat