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My number theory is a bit rusty, so i am trying to recall how to work this problem out. I know that the euler theorem would state that $2^{\phi(101)} \equiv 1$ mod $101$

But in this case, $\phi(101)$ is $100$ so i only know how to calcuate $2^{100}$ mod $101$

Bill Dubuque
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user2327195
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    Can you figure out the residue class $x$ such that $2x\equiv 2^{100}\equiv1\pmod{101}$? – Jyrki Lahtonen May 05 '16 at 05:19
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    Ah ha, yes. So i i can see that i can equate $2*51 \equiv 2^{100}$ mod $101$ thus meaning $2^{1}2^{99} \equiv 2^{100}$ thus $2^{99} \equiv 51$ Thanks a bunch!! – user2327195 May 05 '16 at 05:25
  • By Fermat or mod order reduction (1st dupe) $!\bmod 101!:\ 2^{99}\equiv \frac{1}2\equiv \frac{102}{2}\equiv 51,$ by twiddling or mod halving (cf. 2nd dupe). – Bill Dubuque Mar 26 '24 at 23:45

2 Answers2

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Note that $$2^9=512 \equiv 7 \pmod {101}$$

So we have that $$2^{99}\equiv 7^{11} \pmod {101}$$

Can you do it now?

SchrodingersCat
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Mar 26 '24 at 23:47
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By Euler's Theorem, we can state that 2^(101) ≡ 1 mod 101. Since (101) = 100, we have 2^100 ≡ 1 mod 101. Then, 2 2^99 ≡ 1 mod 101, which is same with 2 2^99 ≡ -100 mod 101. Since gcd(2,101) = 1, we can divide both side by 2. Therefore, we get 2^99 ≡ -50 mod 101, which is same with 2^99 ≡ 51 mod 101. Therefore, 2^99 ≡ 51 mod 101.

Mr. Duysex
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