Fool's problem of the day, Find the remainder when \( 44^8 \) is divided by \( 119 \). PS:This problem is originated from one of the myininaya's reply at OS feedback chat.

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Fool's problem of the day, Find the remainder when \( 44^8 \) is divided by \( 119 \). PS:This problem is originated from one of the myininaya's reply at OS feedback chat.

Mathematics
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eh... i hate these
67 by brute force: http://fwd4.me/0lMq
hehe

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For myin, the constraint is higher \( 44^{86} \) ;-)
Its extremely easy. Here: 44^2 = 1936 = 32 mod 119 implies 44^4 = 32^2 = 1024 = 72 mod 119 implies 44^8 = 72^2 = 5184 = 67 mod 119
Sure Aron, now try the one with higher constraints.
Well, the second problem can be reduced to solving for x in\[44^{10} x \equiv 1 \mod \; 119\]using Euler's totient function. From there, we can find that\[x=60\]Unfortunately, I still need the help of Wolfram for that last step :(
Okay,here it is : 86 = 2*43., 44 = 4*11 44^8 = 67 mod 119 implies 44^40 = 67^ 5 = 16 mod 119 implies 44^43 = 64*16= 72 mod 119 implies 44^86 = 72^2 = 5184 = 67 mod 119.Done!
Aron, you are right apart from the fact the the remainder is not 67.
Alternatively, we know that \[44^{86}=44^{64}*44^{16}*44^4*44^2\]and by computing successive squares of 44 modulo 119 we can also find that \[44^{86} \equiv 60 \mod \: 119\]
This is the the same method Aron used for finding \[44^8 \mod \: 119\]
That's right KIngGeorge, however it's extremely tedious when you don't have any electronic help.
Is there a way to do it quickly without electronic help?
There always is :)
You can use Euler's & Fermat's Theorem!
How so?
@Fool I am sorry! I just checked the remainder to be 60.

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