I have that $2^{947} (\bmod 1373)$ how does one solve this without a calculator? Can you separate it into nice $2^x2^y$ or $(2^x)^y$? I'm really not sure how to go about this problem. Thanks for any help!
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In their question, b<c, but my c<b and my c is prime. With this I can't really divide out my b or factor c. And I won't be able to find a cycle of $2^x$ mod 1373 – Sarah Oct 22 '19 at 00:00
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1If you check the answers for that question, you'll see they describe various techniques which cover basically all cases, including yours. For example, one which works quite well in general, and can work for you too, is this answer. – John Omielan Oct 22 '19 at 00:03
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General strategy: You never need large numbers if you reduce as you go along. Since $$ 2^{10} = 1024 \equiv -349 \pmod {1373} $$ you have a good start. Squaring $349$ will tell you $2^{20}$, and so on.
John Omielan
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Ethan Bolker
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and reducing exponent mod $p-1$ using negative values you have it congruent to $2^{-425}$ – Oct 22 '19 at 01:13
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Wouldn't this take forever to get to 947? And I tried with the -425 but couldn't get an answer – Sarah Oct 22 '19 at 02:29
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It won't take forever since you square each time. So the exponents are $10, 20, 40, 80, 160, \ldots$. You can combine those powers (and the first $9$) bu expressing $1347$ in binary. That's essentially the idea behind one of the linked answers). – Ethan Bolker Oct 22 '19 at 02:35