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RolyPoly
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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?
 one year ago
 one year ago
RolyPoly Group Title
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?
 one year ago
 one year ago

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RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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\(91^{70}\)?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
there the same thing, how much modular arithmetic do you know
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I only know how to find x^y mod (n) using the trick that express y in binary number first, then take mod.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Why are you deleting your comment?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
oh your refering to the binary exponentiation algorithm
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
This one: http://openstudy.com/updates/51165c4de4b09e16c5c86007
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$91^{71}\equiv 91 \text{ mod 100}$$ is what you are trying to show
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Suppose the question is find the last two digits of \(91^{71}\). How would you start?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
Do you know what the mod operator does?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
if I were to write 91^71 in base 10
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
it would be $$91^{71}=a+b10+c10^2+d10^3...$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
now we can reduce that modulo 100, ok
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Ehh!!! So, finding the last two digits of n is the same as finding n mod 100?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
so $$91^{71}\text{ mod 100}=a+10b$$ Where a is are first digit and b are second digit
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
yes pretty much
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
And finding the last n digits of m is the same as finding m mod (10^n)?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
Which you can compute using your exponentiation by squaring algorithm or using a variety of other techniques
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
though for small numbers like yours, I wouldn't use that technique
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
What technique would you use?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Don't tell me calculator ~_~
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
well for $$91^{71}\text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
I would reduce the 91 modulo 100, so I get,
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$91^{71} \text{ mod 100} = (9)^{71} \text{ mod 100}= 9^{71} \text{ mod 100}$$
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
"reduce the 91 modulo 100"?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
Yes, I think maybe you should read up a little on modular arithmetic,
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$91\equiv 9 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$91^{71}\equiv (9)^{71}\text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
so now we just have to compute 9 to the 71st power modulo 100
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
but $$(1)^{71}=1$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
so its just computing $$9^{71}\text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
Now 9 is coprime to 100
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
and $$\phi(9)=6$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{6}\equiv 1 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
raising both sides to the 12th power, gives
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{72}\equiv 1 \text{ mod 100 }$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{72}\equiv 1 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
but wqe want $$9^{71} \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
if we let $$9^{71}=x$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
we have $$9x\equiv 1 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I'll tag you if I have questions. I need to work it out myself to see if I really understand it. Sleep well :)
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
alright later
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@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)???
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
nice catch, this changes how you would proceed
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
We want $$\phi(100)$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
which is 40
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Don't be so lazy... :(
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Yay!!! Thanks!! :)
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{10}\equiv 1 \text{ mod 100 }$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
Which you can get by exponentiation by squaring
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
nvm im being stupid the original problem was $$9^{71} \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
this is the same as $$3^{142} \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
ahh maybe you should use your algorithm, i am to lazy to figure this out
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
142 is a LARGE number!!!
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
then use your technique
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
use the fact $$9^{10}\equiv 1 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
so $$9^{70}\equiv 1 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{71}\equiv 9\text{ mod 100}$$
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
9^(40) ≡1 mod 100 since phi(100) is 40 o_o
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{71}\equiv 9 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{71}\equiv 91 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$91^{71}\equiv 91 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
the first two digits of $$91^{71}$$ are 1 and 9
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
9^(10) ≡1 mod 100 < How do you work this out?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^2\equiv 19 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^4\equiv 19^2 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^4\equiv 39 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^8 \equiv 39^2 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^8\equiv 21 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{10}\equiv 21*9^2 \text{ mod 100}$$
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
$$9^{10}\equiv 1 \text{ mod 100}$$
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
I raised both sides to the 7th power
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Ok, I didn't notice that, sorry!!
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
If $$a\equiv b \text{ mod m}$$ and $$c\equiv d \text{ mod m}$$ then $$ac\equiv bd \text{ mod m}$$
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

Jack17 Group TitleBest ResponseYou've already chosen the best response.1
ye no problem
 one year ago
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