Sign up ×  How can i find the ones digit for the number
$$2^{98}$$  share|cite|improve this question 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.  share|cite|improve this answer
$2^{1+4}=32=2\pmod{10}$ hence $2^{1+4n}=2\pmod{10}$ for every
$n\geqslant0$ hence $2^{1+4\cdot24}=2^{97}=2\pmod{10}$ hence
$2^{98}=2\cdot2^{97}=2\cdot2=4\pmod{10}$.  share|cite|improve
this answer  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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