NumbersThe expression 2^6n – 4^2n , where n is a natural number is always divisible by
Question
Numbers
The expression 2^6n – 4^2n
, where n is a natural number, is always divisible by...
Solution
To determine if the expression 2^6n - 4^2n is always divisible by a natural number, we can break it down step by step.
Step 1: Simplify the expression. 2^6n can be written as (2^6)^n, which is equal to 64^n. Similarly, 4^2n can be written as (4^2)^n, which is equal to 16^n.
So, the expression becomes 64^n - 16^n.
Step 2: Factor out common terms. We can factor out 16^n from both terms: 64^n - 16^n = (4^3)^n - (4^2)^n = (4^3 - 4^2)^n
Step 3: Simplify the expression further. 4^3 - 4^2 = 64 - 16 = 48
So, the expression becomes 48^n.
Step 4: Conclusion. Since 48^n is a multiple of 48 for any natural number n, we can conclude that the expression 2^6n - 4^2n is always divisible by a natural number.
Similar Questions
The expression 2^6n – 4^2n , where n is a natural number is always divisible byOptions15364818
Choose the correct option.When n is divided by 6, the remainder is 4. When 2n is divided by 6, the remainder is
Using the digits 4, 6, 3 and 8, form the greatest 4-digit number that is divisible by 2.Fill in the blank with numerals.Check Answer
Create an expression that shows "eight divided by two raised to the power of four".Then, evaluate the expression.Reset Submit
An even number can be expressed as the square of an integer as well as a cube of another integer. Then the number has to be necessarily divisible by:
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.