Question

I made a challenge for math lovers. Who answers correctly deserves a brainliest. The question is step-by-step. ( I ) Show that xⁿ-1 is always divisible by x-1. ( II ) Prove the divisibility test of 9. ( III ) Prove the divisibility test of 7 in octal. ( IV ) Prove the divisibility test of n-1, in base n.

Answer 1

Answer:

answer of 1st one

1/3×9

3 it shows 9 is divisible by 3

Answer 2

Let's do the divisibility test of nine first. It will suggest a way to solve the problem.

( II ) Divisibility test of 9

A good example of this test is 100a+10b+c.

$100a+10b+c=(99a+9b)+a+b+c$

Inside the bracket is divisible by nine, so finding a sum of every digit becomes the divisibility test.

From this ( II ), we may approach a way of regrouping numbers.

( I ) Divisibility of xⁿ-1

What is one raised by another number? It is multiplying one to itself, by that times.

From above, we know that 1ⁿ=1 and we also know that 1ⁿ-1=0. We can conclude x-1 is a factor of the given polynomial.

( III ) Divisibility test in octal

From ( I ), we have proved xⁿ-1 is divisible by x-1. It is actually the same for numbers, then 8ⁿ-1 is also divisible by 7.

Every digits in octal has a digit value of 8ⁿ.

We can approach in a same way like ( I ).

$8^2a+8b+c=7Q+(a+b+c)$

Regrouping the numbers, we know a+b+c is the remainder of division by 7.

( IV ) Divisibility test of x-1

Used x instead because ⁿ was the only superscript.

Every digit in base k has a digit value of xⁿ. From ( I ), we have proved that xⁿ-1 is divisible by x-1. From ( III ), we know that sum of every digits will become the remainder of division by x-1. QED