Question

A number of the form 8n
, where n is a natural number greater than 1, cannot be
divisible by __

Answer 1

Answer:

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Step-by-step explanation:

A number 8n

We need to check whether it can end with zero

Solution:

We know that any number is multiplied by 5 or 10 or by the multiples of 10 ends with zero.

So here the number 8 has factors 1,2,4 and 8.

It does not contain 5 and 10 as prime factors

8n = (2 X 4)n doesn’t have 5 in its prime factorization

Hence, 8n cannot end with the digit 0

Answer 2

A number of the form 8ⁿ , where n is a natural number greater than 1, cannot be divisible by 40

Given:

• A number of the form 8ⁿ
• n is a natural number greater than 1

To Find:

• Number by which 8ⁿ is not divisible
• A. 1
• B. 40
• C. 64
• D. 2²ⁿ

Solution:

"Prime Factorization is finding prime numbers which when multiplied together results in the original number"

Step 1:

Prime factorize 8  and rewrite 8ⁿ

8 = 2 x 2 x 2

8ⁿ = (2 x 2 x 2)ⁿ

Step 2:

Rewrite (2 x 2 x 2)ⁿ  using (a x b)ⁿ = aⁿ x bⁿ   and  xᵃ * xᵇ = xᵃ⁺ᵇ

8ⁿ =  2ⁿ x 2ⁿ x 2ⁿ

=> 8ⁿ = 2ⁿ x 2ⁿ⁺ⁿ

=> 8ⁿ = 2ⁿ x 2²ⁿ

Hence Divisible by 2²ⁿ

Step 3:

Taking n =2 as n is greater than 1

8ⁿ = 8² = 64 hence Divisible by 64

Step 4:

Every natural number is divisible by 1

Hence 8ⁿ  Divisible by 1

Step 5:

Prime factorize 40

40 = 2 x 2 x 2 x 5  

There is a factor 5 in 40 which is not there in 8ⁿ  Hence,

8ⁿ cannot be divisible by 40

So correct option is B) 40

A number of the form 8ⁿ , where n is a natural number greater than 1, cannot be divisible by 40

Complete Question is here: https://brainly.in/question/48865210

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