Show that for all positive integers $n\geq 2$ the last digit of the number $2^{2^n}+1$ is $7$ .
Problem
Source:
Tags: number theory, TST, Kosovo, 2016
andrei.pantea
09.01.2017 18:40
The last digit of $2^{4k}$ is $6$.
tarzanjunior
09.01.2017 18:40
Use induction.
rafayaashary1
09.01.2017 18:50
Because $\text{ord}_5(2)=4\mid 2^n\implies 2^{2^{n}}+1\equiv 1\pmod{2},\ 2^{2^{n}}+1\equiv 1+1\equiv 2\pmod{5}\implies 2^{2^{n}}+1\equiv 7\pmod{10}$
claserken
09.01.2017 19:07
Basically, we are asked to prove $2^{2^n}\equiv 6\pmod{10}$. We induct on the base case $n=2$, which is trivially true. Realize then that $2^{2^{n+1}}=\left (2^{2^n}\right )^2$, by induction hypothesis we get that $2^{2^n}\equiv 6\pmod{10}$, so $2^{2^{n+1}}\equiv 36\pmod{10}$ or $2^{2^{n+1}}\equiv 6\pmod{10}$, completing the induction$\blacksquare$
ali3985
17.04.2019 18:14
claserken wrote:
Basically, we are asked to prove $2^{2^n}\equiv 6\pmod{10}$. We induct on the base case $n=2$, which is trivially true. Realize then that $2^{2^{n+1}}=\left (2^{2^n}\right )^2$, by induction hypothesis we get that $2^{2^n}\equiv 6\pmod{10}$, so $2^{2^{n+1}}\equiv 36\pmod{10}$ or $2^{2^{n+1}}\equiv 6\pmod{10}$, completing the induction$\blacksquare$
Bravo.
ubermensch
17.04.2019 18:39
This is a TST problem? Just use the fact that $2^{2^n} = 2^{2^{n-2} \cdot 4} =16^{2^{n-2}}=6^k(mod10)=6(mod 10)$
ali3985
17.04.2019 19:19
ubermensch wrote: This is a TST problem? Just use the fact that $2^{2^n} = 2^{2^{n-2} \cdot 4} =16^{2^{n-2}}=6^k(mod10)=6(mod 10)$ Are you sure $6^k(mod10)=6(mod 10)$
ali3985
17.04.2019 19:20
ubermensch wrote: This is a TST problem? Just use the fact that $2^{2^n} = 2^{2^{n-2} \cdot 4} =16^{2^{n-2}}=6^k(mod10)=6(mod 10)$ Is this a complete solution
ubermensch
17.04.2019 19:43
ali3985 wrote: ubermensch wrote: This is a TST problem? Just use the fact that $2^{2^n} = 2^{2^{n-2} \cdot 4} =16^{2^{n-2}}=6^k(mod10)=6(mod 10)$ Are you sure $6^k(mod10)=6(mod 10)$ Yes, obviously?... direct consequence of the fact that $6 \cdot 6 =6(mod 10)$... but because it is a very short proof anyways, it would be better to "prove" this, but it is pretty trivial nonetheless.
Rizsgtp
27.06.2019 06:04
WeakMathemetician
29.06.2019 12:36
$$2^{2^{n+1}} + 1 \equiv ((2^{2^n}+1)-1)^2+1$$