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I have to find the last two digits of $6543^{210}$, my strategy is to use the Euler theorem and then some algebra to reduce this to $6543^{10}$, however I can't think of any easy way to proceed after this, any ideas?

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    Related: [How do I compute $a^b\,\bmod c$ by hand?](http://math.stackexchange.com/questions/81228)2015-08-20

3 Answers 3

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One can use the binomial theorem (thrice). To do so, write $h$ for everything that is a multiple of $100$, possibly varying from line to line.

Since $6543=43+h$, one knows that $ 6543^{210}=(43+h)^{210}=43^{210}+h. $ Now, $ 43^{210}=(3+40)^{210}=3^{210}+210\cdot3^{209}\cdot40+h=3^{210}+h. $ Finally, $ 3^{210}=9^{105}=(-1+10)^{105}=-1+105\cdot10+h=1049+h=49+h, $ that is, $6543^{210}=49+$ some multiple of $100$.

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    well,I am just an infant :)2011-11-27
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This might be the fastest way: For the last two digits, you want to look modulo $100$. Notice that your number is relatively prime to $100$, and that $\phi(100)=40$. By Euler's theorem $6543^{210}\equiv 6543^{10}\equiv 43^{10}\pmod{100}.$ Here we can try using repeated squaring. $43^2\equiv 49\pmod{100}.$ $43^4\equiv 49^2\equiv 1\pmod{100}.$ Since $43^4\equiv 1$, we see that $43^{10}\equiv 43^2\equiv 49\pmod{100}.$

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    @MaX Because $6543=6500+43=65*100+43\equiv 43\pmod{100}$.2011-11-28
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Binomial Theorem $\Rightarrow\rm mod\ 100\!:\ (-7\!+\!50)^{\large 2+4N}\!\equiv (-7)^{\large 2+4N}\! \equiv 7^{\large 2}\, $ by $\, 7^{\large 4}\! \equiv (-1\!+\!50)^{\large 2}\!\equiv 1$

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    this is brilliant!2013-05-12