In a correction of an exercise, the teacher simply wrote down
$$2^{107}\text{mod }187 = 161.$$
Is there any way for this to be so easily calculated?
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4Well, $187=11\times 17$ and it is easy enough to work out $2^{107}$ modulo each of those primes. That's a good start. – lulu Jan 21 '18 at 16:28
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See https://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand – Arnaud D. Jan 16 '19 at 16:57
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We have $$2^{10}=1024\equiv89\pmod{187}$$ $$89^2=7921\equiv67\pmod{187}$$ $$67^2=4489\equiv1\pmod{187}$$ so $$2^{107}=2^{100}\cdot2^{7}=(2^{10})^{10}\cdot128\equiv89^{10}\cdot128\equiv67^5\cdot128\equiv67\cdot128=161\pmod{187}$$as desired.
ə̷̶̸͇̘̜́̍͗̂̄︣͟
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use that $$2^{40}\equiv 1 \mod 187$$
Dr. Sonnhard Graubner
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1Would you care to explain how to use that? It doesn' t seem to have immediate connections to the question or the (known) answer. – Jan 21 '18 at 16:31
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see here https://en.wikipedia.org/wiki/Fermat%27s_little_theorem – Dr. Sonnhard Graubner Jan 21 '18 at 16:34
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1Sorry, maybe I didn't make myself clear enough: I wasn't asking about well-known theorems, I asked how to use your remark. – Jan 21 '18 at 16:34
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1I don't have a problem, I asked a question: how can your remark help to get to the known answer? If @Atmos can do $2^{27}$ by mental calculation, that's fine, but most can't. – Jan 21 '18 at 16:38
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hm, i have written the reason above, we use this trick very often! – Dr. Sonnhard Graubner Jan 21 '18 at 16:41
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1Ok, I desist. When you understand something, you call it a device. When you don't, you call it a trick. – Jan 21 '18 at 16:43