Find the last Two Digits of these Powers | IMO Longlist

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I’ll be honest, I just noticed that the powers of 9 (mod 100) go [9, -19, 29, -39, 49, -59, 69, -79, 89, -99], which shows that they have a period of 10. This reduces the problem to showing that 9^9^9 = 9^9 (mod 10), and I then made the same parity argument that was seen in the video. Definitely can see why this didn’t make the shortlist

TheodoreBrown
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Easier way with Euler's generalization of Fermat's little theorem.

It basically says that if the base is coprime to the modulus m, then the exponent works mod φ(m).
You have 4 "levels", so the top level works mod φ(φ(φ(100))) = φ(φ(40)) = φ(16) = 8
And 9 mod 8 is 1, so you end up with 9^9^9^1 = 9^9^9. Done.

f-th
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Simple way is by using Euler's totient

sciencefactswithadamuabuba
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just apply Euler's totient function 3 times, since the multiplicative order mod n always divides phi(n):

phi(100) = 40, phi(40) = 16, phi(16) = 8.
9 = 1 mod 8
9^9 = 9^1 = 9 mod 16
9^9^9 = 9^9 mod 40
9^9^9^9 = 9^9^9 mod 100

stanleydodds
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nice to see different solution...often the author's is one of the most efficient thi s time seems more complex but at least creative. Went with finding the period of 9%100 and exp%10 (parity)...seems like chaining the powers of 9 all have the same %100

bait
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Can you do a video that proves this: If gcd(a, b, c) = 1 then there exists a K such that gcd(a+Kb, c) = 1?
Thank you for the nice videos so far!

matchedimpedance