What is the algebraic proof for the remainder of 11 when dividing by 12?

[Physiona]

I'm currently doing a grade 9 paper, and one of the following questions is tripping me up a little bit:

Prove algebraically that the sum of the squares of any three consecutive odd numbers always leaves a remainder of 11, when divided by 12.

My attempt of the question:
I have labelled 3 consecutive odd numbers as:
2n+1, 2n+3, and 2n+5.
I've squared them:
(2n+1)²: 4n²+4n+1
(2n+3)²: 4n²+12n+9
(2n+5)²: 4n²+20n+25

I've attempted to then add all the squared numbers which led me to the expression of: 12n²+36n+35. From here, I have attempted to divide by 12, which leads me to n²+3n+35/12. I'm not sure how to get the remainder of 11, do I take factors out, or is my working solution wrong? Any guidance? Thank you!

 

[member 587159]

You are almost there:

Notice $12n^2 + 36n + 35 = 12n^2 + 36n + 24 + 11 = 12(n^2 + 3n + 2) +11$

And this proves the claim.

 

[PeroK]

You seem to have done all the hard work and then failed to spot the answer when it's staring you in the face!

 

[Physiona]

You are almost there:

Notice $12n^2 + 36n + 35 = 12n^2 + 36n + 24 + 11 = 12(n^2 + 36n + 2) +11$

And this proves the claim.

Aha, that makes sense! Thank you for the quick reply!

 

[Physiona]

You seem to have done all the hard work and then failed to spot the answer when it's staring you in the face!

Yep, I know! I intend to do that, and it often causes me frustration! I've worked it out now, thank you!

 

[hilbert2]

The long division algorithm for polynomials is quite similar to that of decimal numbers, and the remainder is defined in the same way in both. In a more difficult case, the divisor can also contain an unknown variable (unlike the divisor 12 in this example). Here's a link to more info.

http://www.purplemath.com/modules/polydiv2.htm