Work with Exponential Large Number to find two last digits..

[zerostalk]

Here's the quest:

Find two last digits of \(\displaystyle 11^{2011^{2011^{2012}}}\)

I try with mod 100

\(\displaystyle 11^{2011^{2011^{2012}}}\) mod 100

I know that 11 and 2011 are prime numbers

I do with conventional method and I get that \(\displaystyle 11^{2011} = 11 mod 100\)

But, I don't know what to do next..

I also use wolfram, but did'nt get answer...

I consider that the answer is 11 or 21, but I still don't know how tow count it...

 

[CRGreathouse]

Re: Work with Exponential Large Number to find two last digi

I do with conventional method and I get that \(\displaystyle 11^{2011} = 11 mod 100\)

But, I don't know what to do next.

Well, that step doesn't help, so let's start from the beginning.

What is the period of 11 mod 100? That is, find some k such that 11^k = 1 (mod 100). Then you can reduce the exponent, 2011^(2011^2012) mod that number. To do that you'll want to find the period of 2011 mod k.

 

[zerostalk]

Re: Work with Exponential Large Number to find two last digi

What is the period of 11 mod 100? That is, find some k such that 11^k = 1 (mod 100). Then you can reduce the exponent, 2011^(2011^2012) mod that number. To do that you'll want to find the period of 2011 mod k.

I find k = 0, 10, 20, 30, etc or k = 10p, with p = non-negative integer

so 2011^(2011^2012) mod 10p, is it right?

But, I can't understand next step...

 

[CRGreathouse]

Re: Work with Exponential Large Number to find two last digi

I find k = 0, 10, 20, 30, etc or k = 10p, with p = non-negative integer

so 2011^(2011^2012) mod 10p, is it right?

No. The period is 10, so work mod 10 (not mod 10p).

 

[zerostalk]

Re: Work with Exponential Large Number to find two last digi

Can I use euler like this:

\(\displaystyle \phi (100)= 40

2011^{2011^{2012}} mod 40 = 11^{2011^{2012}} mod 40

\phi (40)= 16

{2011}^{2012} mod 16 = 11^{2012} mod 16 = 1

11^{1} mod 40 = 11

then, 11^{11} mod 100 = 11\)

so, the answer is 11. Is it right?

 

[CRGreathouse]

Re: Work with Exponential Large Number to find two last digi

Yes. Though you could replace 40 with 10 and 16 with 4 if you like -- working with smaller numbers is ofen more convenient. (10 works since 11^10 = 1 mod 100, and 4 is phi(10).)

 

[zerostalk]

Re: Work with Exponential Large Number to find two last digi

Yes. Though you could replace 40 with 10 and 16 with 4 if you like -- working with smaller numbers is ofen more convenient. (10 works since 11^10 = 1 mod 100, and 4 is phi(10).)

Oh, thank you very much.....

I read about euler yesterday n found that it will be useful for doing this. But, of course, I still have to learn about number theory...

 

[ershi]

Re: Work with Exponential Large Number to find two last digi

If the period of 11^n mod 100 is 10,
wouldn't you be able to say
11^2011^2011^2012= 11^(8136771452) mod 10 (multiplying it out)

Then 8136771452 = 2 (mod 10)
and
11^2 = 121 = 21 (mod 100)

So that the last two digits of the number are 21?

 

[CRGreathouse]

Re: Work with Exponential Large Number to find two last digi

How do you get 8136771452?

 

[ershi]

Re: Work with Exponential Large Number to find two last digi

I got 8136771452 by multiplying out the exponent
2011*2011*2012