The last two digits of $13^{1010}$. $13^{1010}$
$13^{\phi(100)} \equiv 1 \mod 100$
$13^{40} \equiv 1 \mod 100$
$(13^{40})^{25} \equiv 1^{25} \mod 100$
$13^{1000} \equiv 1 \mod 100$
$13^{1010} \equiv 13^{10} \mod 100$
That's all I got. I don't know how to proceed from there. I tried with $\phi (200)$ but it doesn't help at all.
 A: Use repeated squaring to evaluate $13^{8}\pmod{100}$. $$13^2\equiv 69 \pmod{100}$$
$$13^4=(13^2)^2\equiv(69)^2\equiv 61 \pmod{100}$$
$$13^8=(13^4)^2\equiv(61)^2\equiv 21 \pmod{100}$$
$$13^{10}=(13^8)\cdot(18^2)\equiv69\cdot21\equiv 49 \pmod{100}$$
A: As $1010\equiv2\pmod4,$ let $4n+2=1010\iff n=252$
As $13^2=170-1=-1+170$
Using Binomial Theorem,
$$13^{4n+2}=(-1+170)^{2n+1}\equiv(-1)^{2n+1}+\binom{2n+1}1(-1)^{2n}170^1\pmod{100}\equiv170(2n+1)-1\equiv40n+69$$
Now, $n=252\implies252\cdot4\equiv8\pmod{10}$
$\implies252\cdot40\equiv8\cdot10\pmod{10\cdot10}$
Can you take it from here?
A: You may compute this by exponentation by squaring.
So $13^2 = 169 \equiv 69 \equiv -31 \mod 100$. 
So $13^4 \equiv (-31)^2=961 \equiv -39 \mod 100$. 
So $13^8 \equiv (-39)^2=1521 \equiv 21 \mod 100$. 
Hence $13^{10} = 13^8 \cdot 13^2  \equiv 69 \cdot 21 = 1449 \equiv 49 \mod 100$
Hence the last digits are a 4 and a 9. 
A: $(10+3)^{1010}\mod 100$
$\equiv 1010×(10^1)×(3^{1009})+(3^{1010})\mod{100}$
$\equiv 3^{1010}\mod 100$
$\equiv 9^{505}\mod 100$
$\equiv (10-1)^{505}\mod 100$
$\equiv 505×(10^1)×(-1^{504})+(-1^{505})\mod 100$
$\equiv 5050-1\mod 100$
$\equiv 49\mod 100$
Hence the last two digits are $49$.
