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-

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] .

Answered by alex041103 last updated on 30/Sep/17

W e a r e g o i n g t o p r o v e t h i s b y i n d u c t i o n .

F o r 3^{0} (n = 0) t h e s t a t e m e n t i s t r u e .

{L e t}^{'} s s u p p p o s e t h a t t h e s t a t e m e n t i s

t r u e f o r 3^{n} a n d 3^{n} = 100 A_{n} + 10 B_{n} + C_{n}

(s t a t e m e n t : B_{n} \equiv 0 (m o d 2))

\Rightarrow 3^{n + 1} = 3 \times 3^{n} = 3 (100 A_{n} + 10 B_{n} + C_{n}) =

= 100 (3 A_{n}) + 10 (3 B_{n}) + (3 C_{n}) =

= 100 A_{n + 1} + 10 B_{n + 1} + C_{n + 1}

{L e t}^{'} s d e f i n e m o d (x, n) a s m o d (x, n) \equiv x (m o d n)

w h e r e m o d (x, n) = [0, n) .

\Rightarrow B_{n + 1} = \frac{m o d ((30 B_{n} + 3 C_{n}), 100) - m o d ((30 B_{n} + 3 C_{n}), 10)}{10}

B u t 30 B_{n} + 3 C_{n} \equiv 3 C_{n} (m o d 10)

\Rightarrow B_{n + 1} \equiv 3 B_{n} + (t e n s o f 3 C_{n}) (m o d 2)

C_{n} c a n b e 1, 3, 7, 9 a n d 3 C_{n} c a n b e

3, 9, 21, 27 \Rightarrow t e n s o f 3 C_{n} c a n b e

0, 0, 2, 2 w h i c h a r e a l l e v e n

\Rightarrow B_{n + 1} \equiv 3 B_{n} (m o d 2) \equiv B_{n} (m o d 2)

S o b e c a u s e B_{0} \equiv 0 (m o d 2) \Rightarrow

B_{n} \equiv 0 (m o d 2) f o r \forall n .

S o t h e s t a t e m e n t i s t r u e .

Q . E . D .

Commented by Tinkutara last updated on 01/Oct/17

Thank you very much Sir!