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How can i find the ones digit for the number $$2^{98}$$

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Do you mean the number of ones in the decimal representation of 2^98 –  Amr Nov 15 '12 at 14:29
Hint : find $mod 10$ –  Theorem Nov 15 '12 at 14:29

3 Answers 3

up vote 4 down vote accepted


$2^6=8^2 = 64$.

$2^{12} = 64^2 = \ldots6$.

$2^{24} = (\ldots6)^2 = \ldots6$.

$2^{48} = (\ldots6)^2 = \ldots6$.

$2^{49} = 2\cdot(\ldots6) = \ldots2$.

$2^{98} = (\ldots2)^2 = \ldots 4$.

So the answer is 4.

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$2^{1+4}=32=2\pmod{10}$ hence $2^{1+4n}=2\pmod{10}$ for every $n\geqslant0$ hence $2^{1+4\cdot24}=2^{97}=2\pmod{10}$ hence $2^{98}=2\cdot2^{97}=2\cdot2=4\pmod{10}$.

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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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