prove-that-6-n-6-mod-10-for-any-n-Z-

Question Number 76228 by arkanmath7@gmail.com last updated on 25/Dec/19

p r o v e t h a t 6^{n} \equiv 6 (m o d 10), f o r a n y n \in Z^{+}

Commented by turbo msup by abdo last updated on 25/Dec/19

\Leftrightarrow 6^{n} - 6 \equiv 0 [10] \Leftrightarrow 10 d i v i d e 6^{n} - 6

l e t x_{n} = 6^{n} - 6 b y r e c u r r e n c e

n = 1 x_{1} = 6 - 6 = 0 (t r u e)

l e t s u p p o s e x_{n} d i v i d e d b y 10 a n d

x_{n + 1} = 6^{n + 1} - 6 = 6^{n} . 6 - 6

x_{n} = 10 k \Rightarrow x_{n + 1} = (10 k + 6) 6 - 6

= 6 \times 10 k + 30 = 10 (6 k + 3) \equiv 0 [20]

Commented by abdomathmax last updated on 25/Dec/19

x_{n + 1} \equiv 0 [10]

Answered by Rio Michael last updated on 25/Dec/19

L e t s p r o v e b y i n d u c t i o n .

p r o v e f o r n = 1, 6^{1} \equiv 6 (m o d 10)

a s 6 - 6 = 0 ∣ 10

A s u m e f o r n = k \Rightarrow

6^{k} \equiv 6 (m o d 10) \Rightarrow 10 ∣ (6^{k} - 6)

10 ∣ (6^{k} - 6) \Rightarrow \exists m \in Z : 6^{k} - 6 = 10 m

p r o v e f o r n = k + 1

6^{k + 1} \equiv 6^{k} . 6^{1} \equiv 6 (m o d 10)

6^{k} = 10 m + 6

6^{k + 1} \equiv (10 m + 6) 6 w h i c h i s a m u l t i p l e o f 6

h e n c e i t i s t r u e f o r n = k + 1

t h u s \forall n \in Z^{+} 6^{n} \equiv 6 (m o d 10)

Commented by arkanmath7@gmail.com last updated on 25/Dec/19

t h n x