Another two of 'em

Days ago I gave an easy and medium problem from a book of math puzzles. Here are two more, an easy and a difficult.

III.
For every triple (a, b, c) of non-zero real numbers, form the number
a/|a| + b/|b| + c/|c| + abc/|abc|.
(I am using my keyboard's | symbol to denote absolute value.)

The set of all numbers formed is what?


IV.
Let g(x) = x^5 + x^4 + x^3 + x^2 + x + 1. What is the remainder (as a constant number or some polynomial function of x, whichever the case may turn out to be) when the polynomial g(x^12) is divided by the polynomial g(x)?

(Here I use my keyboard's ^ symbol to denote a power.)

III. Looks like Reveal Hidden Content

Reveal Hidden Content

The book agrees with talzamir's answer to III.

Talzamir is still working on IV.

IV.
Let g(x) = x^5 + x^4 + x^3 + x^2 + x + 1. What is the remainder (as a constant number or some polynomial function of x, whichever the case may turn out to be) when the polynomial g(x^12) is divided by the polynomial g(x)?

6

Originally posted by iamatiger

iamatiger's answer of 6 matches the book.

The book gives two different lines of derivation. One starts with using the identity
(x-1)(x^n + x^(n-1) + .. + x + 1) = x^(n+1) - 1,

and ends with
g(x^12) = (x-1) g(x) P(x) + 6.


The other starts with
g(x^12) = g(x) Q(x) + R(x),
where Q is a polynomial and R is the remainder being sought. It makes use of the complex sixth roots of unity.

Do either of those sound like your route to the answer?

Not really the same as either of those, after trying long division I noticed that

(x^(k) - x^(k-1)) * (x^5 + x^4 + x^3 + x^2 + x + 1) = x^(k+5) - x^(k-1)

carrying that 1 step further
(x^(k) - x^(k-1) + x^(k-6) - x^(k-7) )*(x^5 + x^4 + x^3 + x^2 + x +1)=
x^(k+5) - x^(k-7) {equation a}

for the solution we have to divide
x^(5*12) + x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12) + x^(0*12)
by (x^5 + x^4 + x^3 + x^2 + x + 1)

so setting k to 55, we will first have on the top of the long division:
x^55 - x ^54 + x^49 - x^48
{equation a} says this will be equal to x^60 - x^48 = x^(5*12) - x(4*12)
so the remainder, on the bottom of the long division, at this point, will be:
2x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12) + x^(0*12)

by extension of the above we can see the remainder will follow the sequence:
1x^(5+12) + x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12)
2x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12)
3x^(3*12) + x^(2*12) + x^(1*12)
4x^(2*12) + x^(1*12)
5x^(1*12)
6

so we will stop the long division with the remainder 6x^(0*12), i.e. 6

the top of the long division will be the long looking polynomial:
+0
+1(x^55 - x^54 + x^49 - x^48)
+2(x^43 - x^42 + x ^37 - x^36)
+3(x^31 - x^30 + x^25 - x^24)
+4(x^19 - x^18 + x^13 - x^12)
+5(x^7 - x^6 + x - 1)

Lots of work--I am impressed!

What sorts of math classes have you taken, if you don't mind saying?

A level Maths, Degree in Chemistry, I am now a radar systems designer

Originally posted by iamatiger
A level Maths, Degree in Chemistry, I am now a radar systems designer