SR :)) I just fix it =))
–
Rony Phong
·
1 year, 11 months ago

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@Rony Phong
–
Nope. Apply mod 7 followed by Fermat's Little Theorem.
–
Pi Han Goh
·
1 year, 11 months ago

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@Pi Han Goh
–
By using Fermat's Little Theorem, we have :
\[2^{6} \equiv 1 \pmod{7} \Rightarrow 2^{8} = 2 ^{2^{3}} \equiv 4 \pmod{7} \]

So that : \(2^{2^{4}} \equiv 8 \pmod{7}, 2^{2^{5}} \equiv 6 \pmod{7}, 2^{2^{6}} \equiv 2 \pmod{7}, 2^{2^{7}} \equiv 4 \pmod{7},.... \)

We can see that if \(m = 4 \cdot n - 1 \) then \(2^{2^{m}} \equiv 4 \pmod{7} \)

Thus ,
\[2^{2^{2017}} \equiv 6 \pmod{7} \]
... And I get stuck here :(
–
Rony Phong
·
1 year, 11 months ago

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@Rony Phong
–
With tower of exponents, you should be applying Euler's Theorem. This tells us that we should find \( 2^n \pmod{ \phi(7) } \) first.
–
Calvin Lin
Staff
·
1 year, 11 months ago

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@Rony Phong
–
Check your working again. Your entire 3rd line is wrong.
–
Pi Han Goh
·
1 year, 11 months ago

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I think it is prime because \(2^{2^{2017}}+1\) is prime and many primes occur at difference of 2, so due to increased probability I chose it to be prime.
–
Lakshya Sinha
·
1 year, 11 months ago

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@Lakshya Sinha
–
I think \(2^{2^{2017}} + 1 \) is not a prime because it's is a Fermat number,
and until now, there is just 5 Fermat number that are primes is : \(2^{2^{0}} + 1; 2^{2^{1}} + 1;2^{2^{2}} + 1;2^{2^{3}} + 1;2^{2^{4}} + 1\)

I have prove that \(2^{2^{2017}} + 3 \) is not a prime number, but thanks anyway.
–
Rony Phong
·
1 year, 11 months ago

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No.

Last few decimal digits of the solution are.... ...0172573696 <-Last Digit Even != Prime

## Comments

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TopNewestnot a prime number – Ramiel To-ong · 1 year, 11 months ago

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SR :)) I just fix it =)) – Rony Phong · 1 year, 11 months ago

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– Pi Han Goh · 1 year, 11 months ago

Nope. Apply mod 7 followed by Fermat's Little Theorem.Log in to reply

So that : \(2^{2^{4}} \equiv 8 \pmod{7}, 2^{2^{5}} \equiv 6 \pmod{7}, 2^{2^{6}} \equiv 2 \pmod{7}, 2^{2^{7}} \equiv 4 \pmod{7},.... \)

We can see that if \(m = 4 \cdot n - 1 \) then \(2^{2^{m}} \equiv 4 \pmod{7} \)

Thus , \[2^{2^{2017}} \equiv 6 \pmod{7} \] ... And I get stuck here :( – Rony Phong · 1 year, 11 months ago

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Euler's Theorem. This tells us that we should find \( 2^n \pmod{ \phi(7) } \) first. – Calvin Lin Staff · 1 year, 11 months ago

With tower of exponents, you should be applyingLog in to reply

– Pi Han Goh · 1 year, 11 months ago

Check your working again. Your entire 3rd line is wrong.Log in to reply

I think it is prime because \(2^{2^{2017}}+1\) is prime and many primes occur at difference of 2, so due to increased probability I chose it to be prime. – Lakshya Sinha · 1 year, 11 months ago

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I have prove that \(2^{2^{2017}} + 3 \) is not a prime number, but thanks anyway. – Rony Phong · 1 year, 11 months ago

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No.

Last few decimal digits of the solution are.... ...0172573696 <-Last Digit Even != Prime

Approx Solution:

4530133771486397128616413953101758728033979316264652841195935761051588450033937045213076515129395654998815159098974101615963200558154274787748738790211678644410538970061235159505665699096705221172128501364423224257417606736821962911252833066060363075687976750995313953457727475457383546376473713782226055256851957323635233060085750073840982862112794321584114078317329301232817710611834752430415334301714161673057478502612297701175639541114423640398305254720345616205423349569568002063030321637229012200090760969205506795214577344211530377676944078369371486135407432181252977569490258039805559010498267001268 decimal digits

~~ 4.530133771486397×10^606 decimal digits – Jack Bennett · 1 year, 11 months ago

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It is obvious that this number is odd, and hence the last digit is not 6. – Calvin Lin Staff · 1 year, 11 months ago

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