Find the last two digits of $ 7^{81} ?$

Solution 1:

Last $2$ digits of powers of $7$:

$7^1 = 07$

$7^2 = 49$

$7^3 = 43$

$7^4 = 01$

..................................

$7^5 = 07$

$7^6 = 49$

$7^7 = 43$

$7^8 = 01$

So the last two digits repeat in cycles of $4$.

Hence $7^{81}$ has the last two digits same as that of $7^1$. So answer is : $07$

Solution 2:

$7^2=49=50-1$

$\implies 7^4=(7^2)^2=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod {100}$

$$\implies 7^{81}=7\cdot(7^4)^{20}\equiv7\cdot1^{20}\pmod{100}\equiv7$$

In general, Euler's totient theorem or Carmichael's theorem can be used in such cases.

Here

$ \phi(100)=10\cdot\phi(10)=10\cdot\phi(2)\cdot\phi(5)=40$

$\implies 7^{40}\equiv1\pmod{100}$ as $(7,100)=1$

So, $7^{81}=(7^{40})^2\cdot7\equiv1^2\cdot7\pmod{100}\equiv7$

and $\lambda(100)=$lcm$(\lambda(25),\lambda(4))=$lcm$(20,2)=20$

So, $7^{81}=(7^{20})^4\cdot7\equiv1^4\cdot7\pmod{100}\equiv7$

Solution 3:

$7^{81} = 283753509180010707824461062763116716606126555757084586223347181136007$. So, the last two digits are $07$. The first two digits are $28$. In fact, the last three digits are James Bond.

Solution 4:

$\rm{\bf Hint}\ \ mod\,\ \color{#c00}2n\!: \ a\equiv b\, \Rightarrow\, mod\,\ \color{#c00}4n\!:\ \, a^2 \equiv b^2\ \ by \ \ a^2\! = (b\!+\!\color{#c00}2nk)^2\!=b^2\!+\!\color{#c00}4nk(b\!+\!nk)\equiv b^2$

$\rm So,\, \ mod\,\ \color{}50\!:\, 7^{\large 2}\!\equiv -1\Rightarrow mod\ \color{}100\!:\,\color{#0a0}{7^{\large 4} \equiv\, 1}\:\Rightarrow\:7^{\large 1+4n}\equiv\, 7\, \color{#0a0}{(7^{\large 4})}^{\large n} \equiv\, 7 \color{#0a0}{(1)}^{\large n} \equiv 7 $