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Question Number 130727 by EDWIN88 last updated on 28/Jan/21

Commented by MJS_new last updated on 28/Jan/21

$$6$$

Commented by EDWIN88 last updated on 28/Jan/21

$$yes$$

Answered by liberty last updated on 28/Jan/21

$$2^{5\mathrm{k}}=2\left(\mathrm{mod}10\right)\mathrm{\forall }\mathrm{k}\in {\mathbb{Z}}^{+}$$$$320=25\times 12+5\times 4$$$$2^{320}={\left[{\left(2^5\right)}^5\right]}^{12}\times {\left(2^5\right)}^4\left(\mathrm{mod}10\right)$$$$={\left[{\left(2\right)}^5\right]}^2\times \left(2^4\right)\times 2^2\left(\mathrm{mod}10\right)$$$$=2^2\times 2^6=2^5\times 2^3=16\left(\mathrm{mod}10\right)$$$$=6\left(\mathrm{mod}10\right)$$

Commented by JDamian last updated on 29/Jan/21

$$2^{5k}\ne 2\left(mod10\right)$$$$k=2\Rightarrow 2^{10}=1024$$

Answered by mr W last updated on 28/Jan/21

$$2^n=......2ifmod(n,4)=1$$$$2^n=......4ifmod(n,4)=2$$$$2^n=......8ifmod(n,4)=3$$$$2^n=......6ifmod(n,4)=0$$$$sincemod(320,4)=0,2^{320}endswith6$$

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