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 $1000$
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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.
Matt Samuel
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This is formally established via Euler's Totient theorem : $(a,10)=1,\phi(10)=4; a^4\equiv1\pmod{10}$ What if you need to calculate the last two digits ? – lab bhattacharjee Dec 07 '14 at 16:11
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@labbhattacharjee But we don't need to calculate the last two digits, and this seems like an elementary problem so the OP probably wouldn't know the theorem. – Matt Samuel Dec 07 '14 at 16:13
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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?
lab bhattacharjee
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1This seems like overkill to me given the context. It can be computed directly without too much trouble. – Matt Samuel Dec 07 '14 at 16:03
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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 $$9^{1000} \equiv 1 \mod 10$$
Dr. Sonnhard Graubner
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