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Why is the last two digits of \(91^{71}\) the same as 91mod100? Similarly, why is it true that the last two digits of \(81^{81}\) the same as 81mod100, and that of \(71^{91}\) the same as 71mod100?

Discrete Math
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Maybe I should edit my question a bit. It should be "how do I know if the last two digits of \(91^{71}\) is the same as 91mod100? What if I need to find last two digits of\(91^{70}\)?
there the same thing, how much modular arithmetic do you know
I only know how to find x^y mod (n) using the trick that express y in binary number first, then take mod.

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Other answers:

Why are you deleting your comment?
oh your refering to the binary exponentiation algorithm
$$91^{71}\equiv 91 \text{ mod 100}$$ is what you are trying to show
Wait!
ok
Suppose the question is find the last two digits of \(91^{71}\). How would you start?
Do you know what the mod operator does?
if I were to write 91^71 in base 10
it would be $$91^{71}=a+b10+c10^2+d10^3...$$
now we can reduce that modulo 100, ok
Ehh!!! So, finding the last two digits of n is the same as finding n mod 100?
so $$91^{71}\text{ mod 100}=a+10b$$ Where a is are first digit and b are second digit
yes pretty much
And finding the last n digits of m is the same as finding m mod (10^n)?
yes
Which you can compute using your exponentiation by squaring algorithm or using a variety of other techniques
though for small numbers like yours, I wouldn't use that technique
What technique would you use?
Don't tell me calculator ~_~
well for $$91^{71}\text{ mod 100}$$
I would reduce the 91 modulo 100, so I get,
$$91^{71} \text{ mod 100} = (-9)^{71} \text{ mod 100}= -9^{71} \text{ mod 100}$$
"reduce the 91 modulo 100"?
Yes, I think maybe you should read up a little on modular arithmetic,
$$91\equiv -9 \text{ mod 100}$$
therefore
$$91^{71}\equiv (-9)^{71}\text{ mod 100}$$
so now we just have to compute -9 to the 71st power modulo 100
but $$(-1)^{71}=-1$$
so its just computing $$-9^{71}\text{ mod 100}$$
Now 9 is coprime to 100
and $$\phi(9)=6$$
$$9^{6}\equiv 1 \text{ mod 100}$$
raising both sides to the 12th power, gives
$$9^{72}\equiv 1 \text{ mod 100 }$$
$$-9^{72}\equiv -1 \text{ mod 100}$$
but wqe want $$-9^{71} \text{ mod 100}$$
if we let $$-9^{71}=x$$
we have $$9x\equiv -1 \text{ mod 100}$$
now all we have to do is solve for x modulo 100, and we have are solution, I know this looked like it took a while but its because I had to type over this chat, also there is many ways to solve somthing like this, if your just using a computer an algorithm like yours might be better.
But if your doing it by hand, its helpful to be familear with some of this stuff, because sometimes there are 'easyier' ways to do things.
I'll tag you if I have questions. I need to work it out myself to see if I really understand it. Sleep well :)
alright later
@Jack17 Problem problem problem "Now 9 is coprime to 100" "and ϕ(9)=6" " \(9^6≡1 mod 100\)" But isn't it \(a^{\phi(n)} ≡ 1 (\mod n)\) ? Why would we find ϕ(9)???
nice catch, this changes how you would proceed
We want $$\phi(100)$$
which is 40
Don't be so lazy... :(
Yay!!! Thanks!! :)
$$9^{10}\equiv 1 \text{ mod 100 }$$
Which you can get by exponentiation by squaring
nvm im being stupid the original problem was $$-9^{71} \text{ mod 100}$$
this is the same as $$-3^{142} \text{ mod 100}$$
ahh maybe you should use your algorithm, i am to lazy to figure this out
142 is a LARGE number!!!
then use your technique
use the fact $$9^{10}\equiv 1 \text{ mod 100}$$
so $$9^{70}\equiv 1 \text{ mod 100}$$
$$9^{71}\equiv 9\text{ mod 100}$$
9^(40) ≡1 mod 100 since phi(100) is 40 o_o
$$-9^{71}\equiv -9 \text{ mod 100}$$
$$-9^{71}\equiv 91 \text{ mod 100}$$
$$91^{71}\equiv 91 \text{ mod 100}$$
the first two digits of $$91^{71}$$ are 1 and 9
9^(10) ≡1 mod 100 <- How do you work this out?
$$9^2\equiv -19 \text{ mod 100}$$
$$9^4\equiv 19^2 \text{ mod 100}$$
$$9^4\equiv -39 \text{ mod 100}$$
$$9^8 \equiv 39^2 \text{ mod 100}$$
$$9^8\equiv 21 \text{ mod 100}$$
$$9^{10}\equiv 21*9^2 \text{ mod 100}$$
$$9^{10}\equiv 1 \text{ mod 100}$$
Got it! Have never thought that it works in this way... Another problem is that how you can draw the conclusion "so \( 9^{70}≡1 \mod 100\)" from \(9^{10}≡1 \mod 100\) quickly?
I raised both sides to the 7th power
Ok, I didn't notice that, sorry!!
If $$a\equiv b \text{ mod m}$$ and $$c\equiv d \text{ mod m}$$ then $$ac\equiv bd \text{ mod m}$$
Thanks!!! It looks pretty good now :) Btw, if you want to type LaTeX in the same line as the words, use \( instead of \[ . Thanks again for your help :)
ye no problem

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