Find the least significant digit

Question

Explain why the number $(3^{27}\cdot7^{313}\cdot11^{121})^{1000}$ has 1 as its least significant digit. I know I am supposed to use $\pmod{10}$ but I am not sure how to combine the answers of the individual powers and then raise them to $1000$

Answer 1

HINT:

Observe that $3^{27} \cdot 7^{313} \cdot 11^{121}$ is relatively prime with $10$

Now $\phi(10)=4$ and what does Euler's Totient theorem say?

Answer 2

Notice that $3^3=27\equiv 7\pmod{10}$, $3^4\equiv 21\equiv 1\pmod{10}$. $7^2\equiv 9\pmod{10}$, $7^4\equiv 9^2\equiv 1\pmod{10}$. Also, $11^4\equiv 1^4\equiv 1\pmod{10}$. Now powers distribute over products, so $$(3^{27}7^{313}11^{121})^{1000}=3^{27000}7^{313000}11^{121000}$$ and since the exponents are all multiples of $4$ we have that each term in the product is congruent to $1$ mod $10$, hence so is the product of all of them.

Answer 3

we have $$3^{27}\equiv 7\mod 10$$ $$7^{313} \equiv 7 \mod 10$$ $$11^{121}\equiv 1 \mod 10$$ thus we get $$7^2\equiv 9 \mod 10$$ and $$9^{1000} \equiv 1 \mod 10$$

Answer 4

Hint: $\quad3^2=9=10-1,\qquad7^2=49=50-1=5\cdot10-1,\qquad11=10+1$.