How do I find the last two digits $2012^{2013}$?

Question

How do I find the last two digits 2012²⁰¹³

My teacher said this was simple arithmetics(I still don't see how this is simple).

I thought of using Congruence equation as 2012 is congruent to 2 mod 10...but i can't get 2²⁰¹³

Can anyone help please?

Answer 1

You were right in using $2012\equiv 2\mod 10$. Now, notice that there is a repeating pattern in the last digit when you raise $2$ to integer powers:

$2^1=2$

$2^2=4$

$2^3=8$

$2^4 = 16$

$2^5 = 32$

$2^6=64$

$2^7=128$

$2^8=256$

... and so on...

In general:

$2^{4n+1}\equiv 2 \mod 10$

$2^{4n+2}\equiv 4\mod 10$

$2^{4n+3}\equiv 8\mod 10$

$2^{4n}\equiv 6 \mod 10$

$2013$ is of the form $4n+1$. This is enough information to solve your problem.

Answer 2

We work modulo $100$, so we work modulo $2^2=4$ and $5^2=25$. The given number is of course zero modulo four, so we only need it modulo $25$. The Euler indicator of $25$ is $\frac 45\cdot 25=20$, so $12^{20}=1$ modulo $25$. So $$ 2012^{2013}=12^{2013}=12^{20\cdot 100+13}=(12^{20})^{100}\cdot 12^{13} =12^{13}=22 $$ modulo $25$. Among $22$, $22+25$, $22+50$, $22+75$ the one divisible by four is $22+50=72$. So this is the answer.

Computer check, here sage:

sage: R = Zmod(100)
sage: R(2012)^2013
72

Answer 3

Working in mod $100$, we have $$12^{21} \equiv 12$$

Thus $$2012^{2013}\equiv 12^{2013} \equiv 12^{21\times 95+18} \equiv 12^{95} \times 12^{18} \equiv 12^{113} \equiv 12^{21\times 5+8}\equiv 12^{13}\equiv 72$$

Answer 4

You can do all the computations by hand, using Euler's and the *Chines remainder theorem.

First note that $\varphi(100)=\varphi(4)\varphi(25)=2\cdot 20$,so as $2012\equiv 12\mod 100$, we have $$2012^{2013}\equiv 12^{13}=2^{26}\cdot 3^{12}\mod 100.$$ Let's compute first these prime powers $\bmod 4$ and $25$:

• $3$ has order $2\bmod 4$, so $3^{13}\equiv 3\mod 4.$
• $3^{10}\equiv -1\mod 25$, so $3^{13}=3^{10}\cdot 3^3\equiv -2\mod 25$.
• $2^{26}\equiv 0\mod 4$.
• $2^{26}=2^{20}\cdot 64\equiv-11\mod 25$

Thus we have to solve the system of congruences \begin{cases}12^{13}\equiv \color{red}0\mod 4,\\12^{13}\equiv (-2)(-11)=22\equiv \color{red}{-3}\mod 25. \end{cases} The solutions is easy if we have a Bézout's relation between the moduli. In the present case we can start from $$1\cdot 25-6\cdot 4=1$$ from which we deduce the congruence $$2012^{2013}\equiv \color{red}0\cdot1\cdot 25-(\color{red}{-3})\cdot6\cdot 4 = \color{red}{72}\mod 100.$$

Answer 5

Hint: Use congruences mod $100$. You'll find that $12^{21} \equiv 12 \bmod 100$.

Solution: $$ 2012^{2013} \equiv 12^{2013} = 12^{21\cdot 95+18} = (12^{21})^{95} 12^{18} \equiv 12^{95} 12^{18} = 12^{113} = 12^{21\cdot5+8} \equiv 12^5 12^{8} = 2^{13} \equiv 72 \bmod 100 $$

Answer 6

$(10 +2)^k = 10^k + 10^{k-1}*2k + 10^{k-2}{k\choose 2}2^2+.... + 10*2^{k-1}*k + 2^k$.

So the last two digits are determined by $20130*2^{2012} + 2^{2013}$.

And notice the last two digits of $2^k$ repeat.

How long does it take to repeat? Well the last digit repeats every four. $2\to 4\to 8\to 16\to 32$ is back to $2$ so the last to digits must repeat a multiple of $4$. $2^5 = 32$, $2^9=512$ and $2^13\equiv 12*16\equiv 92\mod 100$. $2^{17}\equiv 92*16 \equiv 72\mod 100$ and $2^{21} \equiv 72*16\equiv 52$ $2^{25}\equiv 52*16\equiv 32 \equiv 2^5$ so it repeats every $20$ iterations.

So $2012^{2013} \equiv 30*2^{12} + 2^{13} \equiv 30*96 + 92 \equiv 80 + 92 \equiv 72\mod 100$.