# Another two of 'em

Paul Dirac II
Posers and Puzzles 17 May '14 17:13
1. 17 May '14 17:13
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.)
2. talzamir
Art, not a Toil
17 May '14 19:57
III. Looks like
zero, 4, -4
on a brief glance.

IV. A cursory look would give
(x^72 - 1) * (x - 1) / ((x^12 -1) * (x^6 - 1))
but this is not a satisfying answer yet, and needs more work. ^^
3. 18 May '14 00:271 edit
The book agrees with talzamir's answer to III.

Talzamir is still working on IV.
4. 18 May '14 20:131 edit
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
5. 18 May '14 23:07
Originally posted by iamatiger
Hope that's right, was worked out on pencil & paper - technique on request
6. 19 May '14 01:311 edit
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.

7. 19 May '14 07:454 edits
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)
8. 20 May '14 01:25
Lots of work--I am impressed!

What sorts of math classes have you taken, if you don't mind saying?
9. 22 May '14 22:461 edit
A level Maths, Degree in Chemistry, I am now a radar systems designer
10. 24 May '14 04:22
Originally posted by iamatiger
A level Maths, Degree in Chemistry, I am now a radar systems designer
I read a book on the history of radar about three years back. That was one of the most fruitful collaborations between the UK and the USA.