If n is a natural number, show that for all values of n, (n^3 + 11n) is divisible by 6.?

Question

The answer given was: "

n^3 + 11n
= n x (n^2 + 11).

Since for every n, the product n x (n^2 + 11) is divisible by 6, n^3 + 11n is divisible by 6."

But why? Can someone explain it to me? If possible, please include a few useful sites related. Thanks.

Answer 1

n^3 + 11n = (n^3 - n) + 12n = n(n^2 - 1) + 12n = n(n - 1)(n +1) + 12n

Now n(n-1)(n+1) is the product of three consecutive integers. Given any three consecutive integers, at least one of them is even, i.e. it is divisible by 2, and exactly one of them is divisible by 3,
so n(n - 1)(n +1) is always divisible by 3 X 2 = 6, no matter what n is. Also, 12n = 6 X 2n is always divisible by 6.
Thus, n(n - 1)(n +1) + 12n = n^3 + 11n is always divisible by 6 for any integer n.

Answer 2

To prove that a number is divisible by 6, we have to prove that it's divisible by both 2 and 3. (n^3 + 11n) = n(n^2 + 11). If n is even, then the product is even too. If n is odd, then n^2 is odd and (n^2)+11 is even. So in either case, the product is always even, and therefore always divisible by 2.

Now we have to do something to prove it's always divisible by 3. Every 3rd number is divisible by 3, so another way to state this is that given three consecutive numbers n-1, n, and n+1, one of them has to be a multiple of 3. If n is a multiple of 3, then so is n(n^2 + 11). If n-1 is a multiple of 3, then let n-1=3k for some integer k. Then n(n^2 + 11) becomes (3k+1)((3k+1)^2 + 11), or (3k+1)((9k^2 + 6k + 1 + 11) = (3k+1)((9k^2 + 6k + 12) = (3k+1)*3*(3k^2 + 2k + 4), and thus is a multiple of 3. Finally, if n+1 is a multiple of 3, then let n+1=3k. Then following the same procedure as before we get (3k-1)(9k - 6k + 12), or (3k-1)*3*(3k - 2k + 4). This proves that for all n, the expression n^3 + 11 is always divisible by both 2 and 3, and thus is always divisible by 6.

Answer 3

We want to prove that

n^3 + 11n is divisible by 6 for all natural numbers.

We prove this by mathematical induction.

1) Let n = 1. Then 1^3 + 11(1) = 1 + 11 = 12, which is obviously divisible by 6.

2) Assume that the formula holds true for up to n = k. That is,
assume k^3 + 11k is divisible by 6. (We want to prove that
(k + 1)^3 + 11(k + 1) is divisible by 6).

But,

(k + 1)^3 + 11(k + 1) = (k + 1) [(k + 1)^2 + 11]
= (k + 1) [k^2 + 2k + 1 + 11]
= (k + 1) [k^2 + 2k + 12]
= k^3 + 2k^2 + 12k + k^2 + 2k + 12
= k^3 + 3k^2 + 14k + 12

At this point, I'm going to split up 14k into 11k and 3k.

= k^3 + 3k^2 + 11k + 3k + 12

Now, I'm going to rearrange this such that the 11k is next to the k^3.

= k^3 + 11k + 3k^2 + 3k + 12

Now, I'm going to factor 3 out of the last three terms.

= k^3 + 11k + 3(k^2 + k + 4)
= k^2 + 11k + 3( k(k + 1) + 4 )

= k^2 + 11k + 3k(k + 1) + 12

Let's split this up into three parts by putting the three parts in brackets.

= [k^2 + 11k] + [3k(k + 1)] + 12

k^2 + 11k is divisible by 6, because that is our induction hypothesis.

3k(k + 1) is divisible by 6 because k(k + 1) is an even number.

12 is obviously divisible by 6.

Therefore, we're adding up two values which are divisible by 6. Therefore, the whole thing is divisible by 6.

Which means (k + 1)^3 + 11(k + 1) is divisible by 6.

Therefore, the formula holds true for n = k + 1, and by the principle of mathematical induction,

n^3 + 11n is divisible by 6 for all natural numbers n.

Answer 4

I'm sure you can do a proof by induction.

As to why, let's go through the cases:
- n divisible by 6 - trivial
- n divisible by 2, not divisible by 6:
-- n is 1 mod 3 => n^2 + 11 is 0 mod 3 - done
-- n is 2 mod 3, similar
- n divisible by 3, not divisible by 6
-- n^2 + 11 is 0 mode 2 - done.

Answer 5

In volume theory, a volume a is divisible by technique of a volume b if a / b is an finished volume. on the grounds that 3/8 isn't an finished volume we are saying that 3 isn't divisible by technique of 8. subsequently, n=3 _does_ artwork because 3^2 - a million = 8, and eight is divisible by technique of 8. What become probable stated at school is that n=2 does no longer artwork because 2^2 - a million= 3, and three isn't divisible by technique of 8. therefore n=2 is a counterexample.

Answer 6

Huh?

The values of n are 1, 2, 3, 4.... etc.