BenjaminEE
Modular arithmetic.
21^20 = 1 (mod 100)
From here it's simple to get that
21^(20+1) = 21 (mod 100)
But what about
21^(20-1) = ? (mod 100)
Is there a simple method for this?
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ParthKohli
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\[21^{15} \cdot 21^4 \equiv 1 \cdot 81 \equiv 81 \pmod{100} \]
BenjaminEE
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Thanks, but I'm looking for something more general. Like:
\[p^ {n+1} = p \mod 100\]
Where p^n = 1 mod 100
cahit
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\[21==21\]
ParthKohli
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You have no way but to use the modulos.
cahit
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\[21^2==41\]
cahit
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\[21^3==61\]
ParthKohli
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\[21^4 \equiv 81\]
ParthKohli
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Yes, that is what I used to determine all the mods: pattern recognition. :-)
cahit
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no until you reach 1
later on you can get the powr of 1
ParthKohli
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And write it in some crappy notation so that people think you are smart.
ParthKohli
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Repeats every four terms.
cahit
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\[21^5=1\]
BenjaminEE
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Well what if I have something a bigger repetition pattern? I don't want to brute force the problem.
ParthKohli
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You just have to find a power which is \(1\) or \(-1\) that thing. (thanks @terenzreignz!)
ParthKohli
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\(1\) or \(-1\) modulo*
ParthKohli
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You always have Fermat's Little Theorem!
terenzreignz
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Cute :)
Then again, you could also express it as
(20 + 1)^20 (mod 100)
If that's any easier :D
ParthKohli
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\[a^{b - 1} \equiv 1 \pmod{b}\]