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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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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By the Binomial Theorem, $\rm\ mod\ 100\!:\ (-7+50)^{2+4\:n}\equiv (-7)^{2+4\:n} \equiv 7^2\ $ by $\ 7^4 \equiv (50-1)^2\equiv 1$ |
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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 $\phi(100)=40$. Hence $$6543^{210}\equiv 6543^{10}\equiv 43^{10}\pmod{100}.$$ 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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