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Prove that for all whole values of n the value of the expression (n-3)(n+2)-(n-3)(n+8) is divisible by 6.

User CreativePS
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2 Answers

5 votes

Answer:

Hence, n 3 −n=n(n+1)(n−1) is divisible by 6.

Explanation:

The condition for any number to be divisible by 6 is that the number must be individually divisible by 3 and 2.

Check whether n 3−n is divisible by 3.

n3−n=n(n+1)(n−1)

When a number is divided by 3 then by the remainder theorem, the remainder obtained is either 0 or 1 or 2.

n=3p or n=3p+1 or n=3p+2, where p is some integer.

If n=3p, then the number is divisible by 3.

If n=3p+1, then n−1=3p+1−1=3p. The number is divisible by 3.

If n=3p+2, then n+1=3p+2+1=3(p+1). The number is divisible by 3.

So, any number in the form of n

3

−n=n(n+1)(n−1) is divisible by 3.

Check whether n

3

−n is divisible by 2.

When a number is divided by 2, the remainder obtained is either 0 or 1 by the remainder theorem.

n=2p or n=2p+1, where p is some integer.

If n=2p, then the number is divisible by 2.

If n=2p+1 then n−1=2p+1−1=2p. The number is divisible by 2.

So, any number in the form of n

3

−n=n(n+1)(n−1) is divisible by 2.

Since, the given number n

3

−n=n(n+1)(n−1) is divisible by both 3 and 2. Therefore, according to the divisibility rule of 6, the given number is divisible by 6.

User Rahul Rathore
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8.1k points
3 votes

Answer:

Explanation:

Let's simplify the equation first:

(n-3)(n+2)-(n-3)(n+8)

= n² - n - 6 - (n² + 5n - 24)

= n² - n - 6 - n² - 5n +24

= -6n - 18

Divisible means that the equation can be divided by 6 with no remainder.

If I divide the equation by 6, I get (-n-3)

It goes in evenly, therefore it is divisible by 6

User Amin Ya
by
8.5k points

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