Discrete Math proof

[I Love Math]

Hello! I'm having a bit of a problem with one of the proofs that my instructor gave the class for homework and I'm hoping someone here can help. Here it is:

Show that for every integral n, n^3+2n is divisible by 3.

Now, here's my thinking:

I started by letting n be an even number by letting it equal 2k, where k is an integer. I substituted 2k into n^3+2n and got 8k^3+4k.

I can factor that into 4k(2k^2+1).

I didn't feel like I was getting anywhere so I scrapped that and let n be an odd number by letting it equal 2j+1, where j is an integer. I substituted it into n^3+2n and got 8j^3+12j^2+10j+3. I still don't see if this is helping.

I know that from the beginning, I could have factored out the n to get n(n^2+2).

I'm having trouble figuring out what 'makes' a number divisible by three. I know what makes a number even (2k) or odd (2j+1), but we haven't discussed three in class. I know that if a number's digits add up to something divisible by three, the original number is also divisible by three, but am I even thinking along the right lines here?

Also, I have to figure out how many integers between 100 and 1000 are divisible by 7. Do I subtract 100 from 1000 and divide that answer by 7? 900/7 is 128 remainder 4 so does that mean that there are 128 integers between 100 and 1000 that are divisible by 7?

Thanks for the help guys!

 

[soroban]

Hello, I Love Math!

Here's one approach . . .

Show that for every integer n, n^3+2n is divisible by 3.

Every integer \(\displaystyle n\) is of the form: \(\displaystyle \,3k,\;3k\,-\,1,\;\) or\(\displaystyle \,3k\,+1\)
That is, every integer is a multiple of 3, or one less than a multiple of 3, or one more than a multiple of 3.

Now test each of those forms in: \(\displaystyle N\:=\:n^3\,+\,2n \:=\:n(n^2\,+\,2)\)

If \(\displaystyle n\,=\,3k:\;\;N\:=\:3k(9k^2\,+\,2)\) . . . which is a multiple of 3.

If \(\displaystyle n\,=\,3k\,-\,1:\;\;N\:=\3k-1)[(3k-1)^2\,+\,2] \:=\;(3k-1)[9k^2\,-\,6k\,+\,1\,+\,2]\;=\;(3k-1)(9k^2\,-\,6k\,+\,3)\)
. . . . . . . . . . . . . . . . \(\displaystyle =\:3(3k\,-\,1)(3k^2\,-\,2k\,+\,1)\) . . . a multiple of 3

If \(\displaystyle n\,=\,3k\,+\,1:\;\;N\:=\3k+1)[(3k+1)^2\,+2] \:=\3k+1)(9k^2\,+\,6k\,+\.1\,+\,2] \:=\3l+1)(9k^2\,+\,6k\,+\,3)\)
. . . . . . . . . . . . . . . . \(\displaystyle =\:3(3k+1)(3k^2\,+\,2k\,+\,1)\) . . . a multiple of 3


Since each is a multiple of 3, each is divisible by 3.

 

[I Love Math]

Yay! Thanks so much!

 

[Gene]

Or you can write
n^3+2n =
n(n^2+2) =
n((n^2-1)+3) =
n((n+1)(n-1)+3)
one of them (n,n-1,n+1) is a multiple of 3.
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Gene