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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$43^8 \equiv 1 \bmod(100)$. I just found this by trial though. Hence, $6543^{210} \equiv 43^{210} \bmod(100) \equiv 43^{2} \bmod(100) \equiv 49 \bmod(100)$ – Nov 27 '11 at 19:28
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Related: How do I compute $a^b,\bmod c$ by hand? – Bart Michels Aug 20 '15 at 11:48
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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 $100$.
Did
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1@Sivaram, yes, like a toddler anxiously clutching its parent's hand, I put as many powers of 10 in the picture as I could. – Did Nov 27 '11 at 19:43
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2Because it is afraid of falling on the ground/failing to find the last two digits of $6543^{210}$. (And we are all toddlers in the realm of la mathématique, aren't we?) – Did Nov 27 '11 at 19:58
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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 $$43^{10}\equiv 43^2\equiv 49\pmod{100}.$$
Takahiro Waki
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Eric Naslund
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1I knew this,this is rather the straight forward approach, but don't you think it's cumbersome for a 1 mint rule problem. – Quixotic Nov 27 '11 at 19:35
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@MaX: Not really, repeated squaring is the fastest algorithmic way to do it. – Eric Naslund Nov 27 '11 at 19:39
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btw, in my actual solution I was taking $6543^{210}\equiv 6543^{10}\equiv 3^{10}\pmod{100}$ and then doing the process... it gave correct answer, was it the right approach? – Quixotic Nov 27 '11 at 21:26
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@MaX: Depends. How did you justify $6543^{10}\equiv 3^{10}$? If you gave a reason for this then it is fine. – Eric Naslund Nov 27 '11 at 21:33
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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 $\, 7^{\large 4}\! \equiv (-1\!+\!50)^{\large 2}\!\equiv 1$
Bill Dubuque
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