Use Euler’s $ϕ$-function to calculate $121^{1002}$ mod $100$

Question

Use Euler’s $ϕ$-function to calculate $121^{1002}$ mod $100$ without a calculator.

I know that $121^{1002} = (11^2)^{1002}$ = $11^{2004}$

Applying $ϕ$-function to $11^{2004}$, we get $ϕ$ = $11^{2004}-11^{2003}$ = $11^{2003}(11-1)$ = $10 \times 11^{2003}$

And then, how do I continue? Any help please?

score 4 · Answer 1 · answered Sep 20 '19 at 19:54

4

Alternatively, using binomial theorem: $$121^{1002}\equiv 11^{2004}\equiv (10+1)^{2004}\equiv {2004\choose 1}\cdot 10+1\equiv 41\pmod{100}.$$

answered Sep 20 '19 at 19:54

farruhota

32,168

https://math.stackexchange.com/questions/813379/what-are-the-last-3-digits-of-3100?rq=1 https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345?rq=1 https://math.stackexchange.com/questions/955355/find-the-last-two-digit-of-33100?rq=1 – lab bhattacharjee Sep 21 '19 at 02:17
This would be my favorite! (+1) (Although it doesn't use the $\varphi(x)$-function and the teacher had demanded this...) – Gottfried Helms Jan 13 '20 at 13:21

score 3 · Answer 2 · edited Jan 13 '20 at 10:22

3

Knowing that $\phi(11^{2004}) = 10*11^{2003}$ won't help you in any way that I can see to calculate what $11^{2004} \pmod {100}$ is.

But Are you familiar with Euler's Theorem that states if $\gcd(a,n) = 1$ then $a^{\phi n}\equiv \pmod {n}$?

So as $\phi(100) =40$ and $\gcd(11,100) = 1$, Euler's Theorem allows us to conclude:

$11^{40} \equiv 1 \pmod {100}$.

Can you take it from there?

$11^{2004} \equiv 11^{40*50 + 4}\pmod {100}$. Can you take it from there?

.

$11^{40*50+4} = (11^{40})^{50}*11^4\equiv 1^{50}*11^4 \pmod {100}$. Can you take it from there?

.

$11^4 = 121^2 = (100 + 21)^2$

edited Jan 13 '20 at 10:22

Martin Sleziak

56,060

answered Sep 20 '19 at 19:47

fleablood

130,341

And how do I continue from your last step? – cbc bc Sep 21 '19 at 14:29
Are you asking me what $21^2 \pmod {100}$ is? – fleablood Sep 21 '19 at 14:36
yes cause I need to get it without a calculator as the prof said no calculators – cbc bc Sep 21 '19 at 14:53
1

You need a calculator to figure out what the last two digits of $21^2$ are??? ...... Okay. The pencil and paper method is $21^2 = 21\times 21= 21\times (20 + 1)$. In one row you write "$21$" as the result of $21\cdot 1$ from the "ones" value. And in the next row you write "$42-$" as the result of $2\cdot 21$ from the "tens" value. You add up to "ones" column and get $1+-=1$ in the "ones" column. You add up the $2+2=4$ in the "tens" column. And then you add up $-+4 = 4$ in the "hundreds" column. The result is $441$ and the last two digits are $41$. – fleablood Sep 21 '19 at 15:22
1

But a better method is $21^2 = (210 + 1)^2 = 2^2100 + 2210 + 1\equiv 2210 + 1\pmod{100}$.... I'm pretty sure your professor is assuming $11^4\pmod {100}$ was not meant to be an issue. You could to $11^4 = (10 + 1)^4 = 10^4 + C_{4,3}10^3 + C_{4,2}10^2 + 410 + 1 \equiv 410 + 1\pmod{100}$. – fleablood Sep 21 '19 at 15:27

score 1 · Answer 3 · answered Sep 20 '19 at 19:32

1

Hint:

You're confusing the number to be exponentiated and the modulus.

We need $\varphi(100)==\varphi(2^2)\varphi(5^2)=40$, so $$ 121^{1002}=11^{2004}\equiv 11^{2004\bmod40}\mod 100. $$

answered Sep 20 '19 at 19:32

Bernard

179,256

I didnt understand this : $11^{2004}\equiv 11^{2004\bmod40}$ – cbc bc Sep 20 '19 at 19:41
Since $x^{40}\equiv 1\mod 100$ for all $x$ coprime to $100$… Is it clear? – Bernard Sep 20 '19 at 19:43
If $a^k \equiv c\pmod n$ then $a^{mk + j} \equiv (a^k)^ma^j \equiv c^ma^j \pmod n$. And if $c=1$ and $a^k\equiv 1$ then $a^{mk +j}=(a^k)^ma^j\equiv 1^ma^j \equiv a^j \pmod n$. – fleablood Sep 20 '19 at 19:51
And $2004 = m40 + j$ for some* integer $m$ and some integer $j; 0 \le j < 40$. – fleablood Sep 20 '19 at 19:52
and so $121^{1002}$ mod $100$ $\equiv$ $11^{2004}$$^{mod 40}$ mod $100$? – cbc bc Sep 21 '19 at 14:10
Yes : $;\equiv 11^4\equiv 21^2=400+41\equiv 41\mod 100$. – Bernard Sep 21 '19 at 15:13

score 1 · Answer 4 · answered Sep 20 '19 at 19:35

By Euler's theorem (a generalization of Fermat's little theorem), if $m\geq 1$ and $\gcd(a,m)=1$, then

$$a^{\phi(m)} \equiv 1 \mod{m}$$

So

$$121^{40}\equiv 1 \mod{100}$$ and raising both sides to the power of 25, we have $$121^{1000} \equiv 1 \mod{100}$$ You should be able to finish from here.

score 1 · Answer 5 · answered Sep 20 '19 at 19:52

1

In this case the $\phi-$function is not really required because the cycle of powers of $21$ is very short.

Indeed $21^r\equiv 01,21,41,61,81\pmod{100}$ for $r=0,1,2,3,4$

So $121^{1002}\equiv 21^{1002}\equiv 21^{(1002 \mod 5)}\equiv 21^2\equiv 41\pmod{100}$

answered Sep 20 '19 at 19:52

zwim

29,833

"because the cycle of powers of 21 is very short." Your definition of "very" differs from mine and your willingness to determine the exact length when you initially have no idea how long or short it is is ... admirable. But $\phi(100)= 40$ is exceedingly easy, and noting $2004 \equiv 4 \pmod {40}$ and calculating $11^4\pmod {100}$ are all together a lot less work than calculating the first five cycles of $21^r$ and one has the advantage of knowing the process will end in success before one begins. – fleablood Sep 20 '19 at 19:58
?? what did I do to deserve so much hate ? – zwim Sep 20 '19 at 20:00
Where do you see any hate? – fleablood Sep 20 '19 at 20:02
Interesting approach @zwim ! – Mike Sep 20 '19 at 21:07

Use Euler’s $ϕ$-function to calculate $121^{1002}$ mod $100$

5 Answers5