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Question Number 21682 by Tinkutara last updated on 30/Sep/17

Prove that the ten′s digit of any power of 3 is even. [e.g. the ten′s digit of 3^6 = 729 is 2].

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ten}'\mathrm{s}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{any}\:\mathrm{power} \\ $$$$\mathrm{of}\:\mathrm{3}\:\mathrm{is}\:\mathrm{even}.\:\left[\mathrm{e}.\mathrm{g}.\:\mathrm{the}\:\mathrm{ten}'\mathrm{s}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{3}^{\mathrm{6}} \:=\right. \\ $$$$\left.\mathrm{729}\:\mathrm{is}\:\mathrm{2}\right]. \\ $$

Answered by alex041103 last updated on 30/Sep/17

We are going to prove this by induction. For 3^0 (n=0) the statement is true. Let′s supppose that the statement is true for 3^n and 3^n =100A_n +10B_n +C_n (statement : B_n ≡0(mod 2)) ⇒3^(n+1) =3×3^n =3(100A_n +10B_n +C_n )= =100(3A_n )+10(3B_n )+(3C_n )= =100A_(n+1) +10B_(n+1) +C_(n+1) Let′s define mod(x, n) as mod(x, n)≡x(mod n) where mod(x, n)=[0, n). ⇒B_(n+1) =((mod((30B_n +3C_n ), 100)−mod((30B_n +3C_n ),10))/(10)) But 30B_n +3C_n ≡3C_n (mod 10) ⇒B_(n+1) ≡3B_n +(tens of 3C_n ) (mod 2) C_n can be 1,3,7,9 and 3C_n can be 3, 9, 21, 27 ⇒ tens of 3C_n can be 0,0,2,2 which are all even ⇒B_(n+1) ≡3B_n (mod 2)≡B_n (mod 2) So because B_0 ≡0(mod 2) ⇒ B_n ≡0(mod 2) for ∀n. So the statement is true. Q.E.D.

$${We}\:{are}\:{going}\:{to}\:{prove}\:{this}\:{by}\:{induction}. \\ $$$${For}\:\mathrm{3}^{\mathrm{0}} \left({n}=\mathrm{0}\right)\:{the}\:{statement}\:{is}\:{true}. \\ $$$${Let}'{s}\:{supppose}\:{that}\:{the}\:{statement}\:{is} \\ $$$${true}\:{for}\:\mathrm{3}^{{n}} \:{and}\:\mathrm{3}^{{n}} =\mathrm{100}{A}_{{n}} +\mathrm{10}{B}_{{n}} +{C}_{{n}} \\ $$$$\left({statement}\::\:{B}_{{n}} \equiv\mathrm{0}\left({mod}\:\mathrm{2}\right)\right) \\ $$$$\Rightarrow\mathrm{3}^{{n}+\mathrm{1}} =\mathrm{3}×\mathrm{3}^{{n}} =\mathrm{3}\left(\mathrm{100}{A}_{{n}} +\mathrm{10}{B}_{{n}} +{C}_{{n}} \right)= \\ $$$$=\mathrm{100}\left(\mathrm{3}{A}_{{n}} \right)+\mathrm{10}\left(\mathrm{3}{B}_{{n}} \right)+\left(\mathrm{3}{C}_{{n}} \right)= \\ $$$$=\mathrm{100}{A}_{{n}+\mathrm{1}} +\mathrm{10}{B}_{{n}+\mathrm{1}} +{C}_{{n}+\mathrm{1}} \\ $$$${Let}'{s}\:{define}\:{mod}\left({x},\:{n}\right)\:{as}\:{mod}\left({x},\:{n}\right)\equiv{x}\left({mod}\:{n}\right) \\ $$$${where}\:{mod}\left({x},\:{n}\right)=\left[\mathrm{0},\:{n}\right). \\ $$$$\Rightarrow{B}_{{n}+\mathrm{1}} =\frac{{mod}\left(\left(\mathrm{30}{B}_{{n}} +\mathrm{3}{C}_{{n}} \right),\:\mathrm{100}\right)−{mod}\left(\left(\mathrm{30}{B}_{{n}} +\mathrm{3}{C}_{{n}} \right),\mathrm{10}\right)}{\mathrm{10}} \\ $$$${But}\:\mathrm{30}{B}_{{n}} +\mathrm{3}{C}_{{n}} \equiv\mathrm{3}{C}_{{n}} \left({mod}\:\mathrm{10}\right) \\ $$$$\Rightarrow{B}_{{n}+\mathrm{1}} \equiv\mathrm{3}{B}_{{n}} +\left({tens}\:{of}\:\mathrm{3}{C}_{{n}} \right)\:\left({mod}\:\mathrm{2}\right) \\ $$$${C}_{{n}} {can}\:{be}\:\mathrm{1},\mathrm{3},\mathrm{7},\mathrm{9}\:{and}\:\mathrm{3}{C}_{{n}} {can}\:{be} \\ $$$$\mathrm{3},\:\mathrm{9},\:\mathrm{21},\:\mathrm{27}\:\Rightarrow\:{tens}\:{of}\:\mathrm{3}{C}_{{n}} \:{can}\:{be} \\ $$$$\mathrm{0},\mathrm{0},\mathrm{2},\mathrm{2}\:{which}\:{are}\:{all}\:{even} \\ $$$$\Rightarrow{B}_{{n}+\mathrm{1}} \equiv\mathrm{3}{B}_{{n}} \left({mod}\:\mathrm{2}\right)\equiv{B}_{{n}} \left({mod}\:\mathrm{2}\right) \\ $$$${So}\:{because}\:{B}_{\mathrm{0}} \equiv\mathrm{0}\left({mod}\:\mathrm{2}\right)\:\Rightarrow \\ $$$${B}_{{n}} \equiv\mathrm{0}\left({mod}\:\mathrm{2}\right)\:{for}\:\forall{n}. \\ $$$${So}\:{the}\:{statement}\:{is}\:{true}. \\ $$$${Q}.{E}.{D}. \\ $$

Commented by Tinkutara last updated on 01/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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