How can i find the ones digit for the number $2^{98}$
Finding the ones digit for $2^{98}$
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algebra-precalculus
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0Hint : find $mod 10$ – 2012-11-15
3 Answers
4
$2^3=8$.
$2^6=8^2 = 64$.
$2^{12} = 64^2 = \ldots6$.
$2^{24} = (\ldots6)^2 = \ldots6$.
$2^{48} = (\ldots6)^2 = \ldots6$.
$2^{49} = 2\cdot(\ldots6) = \ldots2$.
$2^{98} = (\ldots2)^2 = \ldots 4$.
So the answer is 4.
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Hint: Try finding the ones digit of the first few powers. You should see a pattern that you can prove. If you just want the specific answer, a spreadsheet with A1=1, A2=mod(2*A1,10), copy down 97 times gets you there.
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$2^{1+4}=32=2\pmod{10}$ hence $2^{1+4n}=2\pmod{10}$ for every $n\geqslant0$ hence $2^{1+4\cdot24}=2^{97}=2\pmod{10}$ hence $2^{98}=2\cdot2^{97}=2\cdot2=4\pmod{10}$.