I am learning the algebra with remainders. For example: find the remainder of $2^{2011}$ divided by $3$ which is $11$. My problems is when i have lower power number then the divisor for example: $3^{56}$ divided $77$ is confusing for me. Someone help pls...
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How did you compute the remainder of $2^{2011}$ divided by $3$ to be $11$ ?! Though it is technically correct since $11\equiv -1\pmod 3$ – Prasun Biswas Mar 24 '18 at 16:47
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See https://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 – lab bhattacharjee Mar 24 '18 at 17:19
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$$\left(2^{10}\right)^{10}\equiv 24^{10}\equiv (-1)^{10}\equiv 1\pmod{25}$$ and $2^{100}$ is obviously a multiple of $4$. Apply the Chinese remainder theorem and you'll get that the last digits are $76$. Plenty of questions on MSE about the same problem. – Jack D'Aurizio Mar 24 '18 at 17:28
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@JackD'Aurizio: I guess we could avoid using CRT here since there are only 4 choices to check $(1,26,51,76)$ and $76$ is obviously the only one among them which is divisible by $4=2^2$. – Prasun Biswas Mar 24 '18 at 17:38
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@PrasunBiswas: of course. That is just a way for solving the system $n\equiv 1\pmod{25}\wedge n\equiv 0\pmod{4}$ by imposing one constraint at a time and recalling that there is a unique solution in $[0,99]$. I.e. by invoking the CRT :) – Jack D'Aurizio Mar 24 '18 at 17:42
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You can just discard hundreds as you go. $2 \times 64 = 128$, discard a hundred so you just have $28$, and then $2 \times 28 = 56$, $2 \times 56 = 112$ but discard a hundred and then $2 \times 12 = 24$, etc. – Lisa Mar 24 '18 at 21:34
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@ViktorDimitrioski Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 27 '18 at 14:12
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We have $$\begin{align}2^{100}&\equiv(2^{10})^{10}\equiv(1024\times1024)^5\equiv(24\times24)^5\equiv(576\times576)^2\times576\\&\equiv(76\times76)^2\times76\equiv5776^2\times76\equiv76^3\equiv76\pmod{100}\end{align}$$
TheSimpliFire
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Note that $\gcd(2,100)=2\neq 1$, so Euler's theorem doesn't apply here. – Prasun Biswas Mar 24 '18 at 16:51
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You can however use CRT here: $2^{100}\equiv 0\pmod{2^2}$ and the residue modulo $5^2$ can be computed using Euler's theorem (here $\gcd(2,5^2)=1$). Then, combine the results by CRT to get the residue modulo $100$ – Prasun Biswas Mar 24 '18 at 16:55
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@TheSimpliFire, Your revised version is essentially equivalent to gimusi's answer. – Prasun Biswas Mar 24 '18 at 16:56
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what if we find after which interval the last 2 digits of $,n^{th}$ power of 2 mod(100) repeats and then find last 2 digits if $,2^{100}$, would that work?? – The Integrator Mar 24 '18 at 16:57
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I do not understand much of your explanation... but thanks for helping :) – Viktor Dimitrioski Mar 24 '18 at 17:13
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Without any advanced tool simply note that $$2^{10}=1024$$ then we can consider $4^{10}=2^{20}=(2^{10})^2$ that is equivalent to consider $$24^2=576$$
user
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