show-that-8-n-3-n-is-divisible-by-5-for-all-natural-number-

Question Number 148732 by ethiork last updated on 30/Jul/21

s h o w t h a t

8^{n} - 3^{n} i s d i v i s i b l e b y 5 f o r a l l n a t u r a l

n u m b e r .

Answered by Rasheed.Sindhi last updated on 30/Jul/21

5 ∣ (8^{n} - 3^{n})

⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣⌣

∵ 8 \equiv 3 (m o d 5)

∴ 8^{n} \equiv 3^{n} (m o d 5)

Answered by mr W last updated on 30/Jul/21

8^{n} = {(5 + 3)}^{n} = 3^{n} + \underset{k = 1}{\sum^{n}} C_{k}^{n} 3^{n - k} 5^{k}

8^{n} - 3^{n} = \underset{k = 1}{\sum^{n}} C_{k}^{n} 3^{n - k} 5^{k} \equiv 0 m o d 5

Answered by physicstutes last updated on 31/Jul/21

let f (n) = 8^{n} - 3^{n}

f (1) = 8^{1} - 3^{1} = 5 = 5 (1) \Rightarrow true for f (1)

assume f (k) is true \Rightarrow 8^{k} - 3^{k} = 5 n, n \in N

now f (k + 1) = 8^{k + 1} - 3^{k + 1} = 8 (8^{k}) - 3 (3^{k})

f (k + 1) = 8 (5 n + 3^{k}) - 3 (3^{k})

= 5 (8 n) + 5 (3^{k})

= 5 (8 n + 3^{k}) since k \in N, 3^{k} \in N

and so (8 n + 3^{k}) \in N \Rightarrow f (k + 1) = 5 p

p \in N \Rightarrow f (k + 1) is true and so for all

natural numbers the expression f (n)

is divisible by 5 .