Find the last $2$ digits of $9^{100}$.
Well, I know that $9^{100}$ mod $4$ is $1$,but I do not know how to find $9^{100}$ mod $25$ hence I do not know how to find $9^{100}$ mod $100$.
Thanks for any help.
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Hari Ramakrishnan Sudhakar
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Ryan Soh
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Have you calculated the last two digits of $9^n$ for a few $n$? You should find a pattern. A spreadsheet and copy down makes it easy. – Ross Millikan Aug 15 '20 at 04:49
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Good idea, but is there a way to calculate $9^{100}$ mod $25$ directly? – Ryan Soh Aug 15 '20 at 04:50
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5What you find is that the values are periodic, so you just need to find where you are in the period. This would work equally well for $9^{10000}$. If you want directly, Alpha gives the answer as $01$. What more do you want? Have you tried anything? -1 – Ross Millikan Aug 15 '20 at 04:53
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1Does this answer your question? Find the last two digits of $3^{45}$ – lab bhattacharjee Aug 15 '20 at 05:31
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1By Euler's theorem, $9^{20}\equiv1\bmod25$ – J. W. Tanner Aug 16 '20 at 01:41
2 Answers
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Hint:observe that $9^{100}={(10-1)}^{100}$
using binomial theorem this can be written in the form of $100k+1$
Hari Ramakrishnan Sudhakar
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Calculate $9^{100}$ mod $25$.
Using the 'brute-force' mod $25$ technique (no calculator necessary),
$\quad 9^{100} = \bigr(\big({(9^2)}^2\big)^5\bigr)^5 \equiv \bigr(\big({6}^2\big)^5\bigr)^5 \equiv {(11^5)}^5 = {\bigr((11)^2(11)^2(11)\bigr)}^5 \equiv {\bigr((-4)(-4)(11)\bigr)}^5 =$
$\quad \quad {\bigr((16)(11)\bigr)}^5 = {176}^5 \equiv 1^5 \equiv 1$ mod $25$
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