Find remainder when $2^{30}\cdot 3^{20}$ is divided by $7$ without using calculator My question is that what will be the remainder when $2^{30}\cdot 3^{20}$ is divided by $7$.
As it'll be practically non-sense to calculate such a large digit so I think one will have to use the binomial theorem. But I have no idea how to begin with it.
Further I also want to inform all you that I have no knowledge regarding the modular arithmetic (general method that discovered in SE and other sites), so please submit an alternative.
Thanks in advance.
 A: Hint: $2^3\equiv1\ \text{mod }7$ and $3^3\equiv-1\ \text{mod }7$.
Edit: The only thing you need from the modular arithmetic is that $$a\cdot b\ \text{mod }7=(a\ \text{mod }7)(b\ \text{mod }7)\ \text{mod }7,$$
meaning the remainder of a product is the remainder of the product of remainders. This is self-evident as $a\cdot b$ must have the same remainder as $(a-7k)\cdot(b-7l)$.
Since exponentiation is just repeated multiplication, we can write
$$2^{30}\cdot3^{20}=(2^3)^{10}\cdot(3^3)^6\cdot 3^2\equiv1^{10}\cdot(-1)^6\cdot9=9$$
So the remainder is $2$.
A: Hint: Use binomial theorem and then divide by 7.
$2^{30}.3^{20}=2^{20}.3^{20}.2^{10}=1024.6^{20} = 1024 .(7-1)^{20}$
$1024\dot \,(1-7)^{20}=1024[1-^{20}C_1.7+^{20}C_2.7^2+\mathrm{other\ terms}]$  
$=1024-^{20}C_1\dot \,7\dot \,1024+^{20}C_2\dot \,7^2\dot \,1024+\ldots+\mathrm{other\ terms}$
Now divide by $7$.
ADD:
Just to complete the answer I guess,
Since all the other terms except first are divisible by 7 your answer lies in only first term. 
A: This still is modular arithmetic, just not using the language:


*

*$2^6 = 64 = 9·7 + 1$, so
$$2^{30} = (2^6)^5 = (9·7+1)^5.$$

*$3^6 = 9·9·9 = 729 = 728 + 1 = 7·104 + 1$, and $3^2 = 9 = 7 + 2$, so
$$3^{20} = 3^2·3^{18} = 3^2 · (3^6)^3 = (7 + 2) ·(7·104 + 1)^3.$$
A: $2^{30} = 8^{10} = 1\pmod 7$, $3^{20} = 9^{10} = 2^{10} = 2^9\cdot 2 = 1\cdot 2 = 2\pmod 7 \Rightarrow 3^{20}\cdot 2^{30} = 2\pmod 7$
A: Let us first understand what is modular arithmetic.
There is an example, 2 and 7 are unequal, but they are equivalent under (mod 5).
The notation $2 \text{ mod }  5 = 7 \text{ mod } 5$ means that when we divide 2 by 5 then the remainder is 2 which is equal to the remainder when we divide 7 by 5. 
You are required to calculate the remainder when $2^{30}.3^{20}$ is divided by 7.
$$2^{30}.3^{20} \equiv (2^{3})^{10}.(3^{2})^{10} (\text{ mod }7)
\equiv 8^{10}.9^{10} (\text{ mod }7) \equiv 1^{10}.2^{10} (\text{ mod }7)$$
You see that when 8 is divided by 7 we get 1 as remainder and when 9 is divided by 7 we get 2 as remainder. 
$$1^{10}.2^{10} (\text{ mod }7) \equiv (2^3)^3.2 (\text{ mod }7) 
\equiv 1^3.2 (\text{ mod }7) \equiv 2 (\text{ mod }7) = 2 $$
Therefore you get 2 as remainder when you divide $2^{30}.3^{30}$ by 7.
There is lemma on remainders which states:
The sum(product) of any two natural numbers has the same remainder, when divided by n, as the sum(product) of their remainders.
