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What is the last digit of $3^{4^{4 \mathrm{n}}}+1$, where $n$ is a natural number?
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In $3^{\text {n }}$, last digit is 3 , if $n=1,9$ if $n=2,7$ if $n=3$ and 1 if $\mathrm{n}=4$ and it is repeated after than
Given expression is $3^{3^{4 \mathrm{n}}}+1$
Let $x=3^{3^{4 n}}+1=3^{81 n}+1$
$\Rightarrow \mathrm{x}=3^{80 \mathrm{n}} \cdot 3^{\mathrm{n}}+1$
Last digit of $x$ will be decided by $3^{\text {n }}$ since $3^{80 \text { n }}$ has power multiple of 4 . $\operatorname{If} n=1$ last digit is $3+1=4$ $\mathrm{n}=2$ last digit is $3^{2}+1=9+1=10$ So, last digit is zero. $\mathrm{n}=3$ last digit is $3^{3}+1=27+1=28$
last digit is 8 If $n=4$ last digit is $3^{4}+1=81+1=82$
last digit is 2 . So, there is no definite value of last digit.
Given expression is $3^{3^{4 \mathrm{n}}}+1$
Let $x=3^{3^{4 n}}+1=3^{81 n}+1$
$\Rightarrow \mathrm{x}=3^{80 \mathrm{n}} \cdot 3^{\mathrm{n}}+1$
Last digit of $x$ will be decided by $3^{\text {n }}$ since $3^{80 \text { n }}$ has power multiple of 4 . $\operatorname{If} n=1$ last digit is $3+1=4$ $\mathrm{n}=2$ last digit is $3^{2}+1=9+1=10$ So, last digit is zero. $\mathrm{n}=3$ last digit is $3^{3}+1=27+1=28$
last digit is 8 If $n=4$ last digit is $3^{4}+1=81+1=82$
last digit is 2 . So, there is no definite value of last digit.
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