Question

If n is any positive integer greater than 2 and p and q are prime number greater than 6, which of the following is necessarily the common divisor of $(n^7-n)$ and $(p^2-q^2)$

(a) 48
(b) 7
(c) 6
(d) 72

Answer 1

Answer:

The correct answer is (a) 48.

Step-by-step explanation:

We must factorise both equations and search for shared factors in order to determine the common divisor of (n⁷ - n) and (p² - q²).

Let's factorise (n⁷ - n) first, which is as follows:

n⁷ - n = n(n⁶ - 1)

We can further factorise (n⁶ - 1 )by using the difference of squares formula:

n⁶ - 1 = (n³ - 1)(n³ + 1)

In order to factorise n³ – 1, we need to use the difference of cubes formula:

n³ - 1 = (n - 1)(n² + n + 1)

The difference of squares formula will now be used to factorise (p² - q²):

p² - q² = (p - q)(p + q)

We can infer that p and q are prime numbers bigger than 6 from the information provided. Due to the oddness of both p and q, their addition (p + q) is also even. Therefore, both (p - q) and (p + q) can be divided by 2.

Based on the factorizations above, we can see that the common divisor must contain at least one factor of 2. Therefore, the only option that meets this requirement is (a) 48, which is divisible by 2⁴.

Therefore, the correct answer is (a) 48.

Similar Questions:

https://brainly.in/question/42431368

https://brainly.in/question/40356392

#SPJ1