Question 1209161: A student writes the six complex roots of the equation z^6 + 2 = 0 on the blackboard.
At every step, he randomly chooses two numbers a and b from the board, erases them, and replaces them with 3ab − 3a − 3b + 4.
At the end of the fifth step, only one number is left. Find the largest possible value of this number.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 730
Explanation
At first it's probably not obvious, but the trick is to realize that we can factor part of it like so
3ab-3a-3b+4
3ab-3a-3b+3+1
3(ab-a-b+1)+1
3( (ab-a)+(-b+1) )+1
3( a(b-1)-1(b-1) )+1
3(a-1)(b-1)+1
In short, 3ab-3a-3b+4 rewrites to 3(a-1)(b-1)+1
The more general template is 3(first-1)*(second-1)+1
where first,second are the randomly selected roots of z^6+2 = 0.
We do this "factorization" of sorts so that we reduce the number of times a,b show up.
Conveniently many pairs of plus 1's and minus 1's will cancel out, which allows us to scale the formula up in a nice pattern.
Let a,b,c,d,e,f represent the 6 roots of z^6+2 = 0 in any order you like.
If we selected a,b and replaced them with 3(a-1)(b-1)+1 we get this smaller set
3(a-1)(b-1)+1, c,d,e,f
The color coding is to separate the items.
Apply the operation to first = 3(a-1)(b-1)+1 and second = c to get
3(first-1)*(second-1)+1 = 3(3(a-1)(b-1)+1-1)*(c-1)+1 = 9(a-1)(b-1)(c-1)+1
We now have this set
9(a-1)(b-1)(c-1)+1, d,e,f
Then apply the same idea to the first two elements to get
3(first-1)*(second-1)+1 = 3(9(a-1)(b-1)(c-1)+1-1)*(d-1)+1 = 27(a-1)(b-1)(c-1)(d-1)+1
This pattern continues where each new letter introduced to this operation will bring in a (___-1) factor.
Then we triple the coefficient out front.
This is what I meant by "scale up nicely".
Here's the full step-by-step look where we always select the two left-most elements of the set
a,b,c,d,e,f
3(a-1)(b-1)+1, c,d,e,f
9(a-1)(b-1)(c-1)+1, d,e,f
27(a-1)(b-1)(c-1)(d-1)+1, e,f
81(a-1)(b-1)(c-1)(d-1)(e-1)+1, f
243(a-1)(b-1)(c-1)(d-1)(e-1)(f-1)+1
which is the last number on the board.
Note that 243 = 3^(n-1) = 3^(6-1) = 3^5
where n = 6 is the number of roots.
This observation should lead directly to the last number on the board without having to compute the intermediate steps.
However I recommend writing out this scratch work so you can see how I'm getting to that expression.
Roots a through f satisfy z^6+2 = 0.
Roots (a-1) through (f-1) satisfy (w+1)^6+2 = 0, since we have z = w+1 or w = z-1.
Plug w = z-1 into (w+1)^6+2 = 0 so you get z^6+2 = 0 again.
One of Vieta's formulas says that the roots of the polynomial multiply to the constant term, when we have an even degree polynomial and the leading coefficient is 1.
(w+1)^6+2 will expand to some messy polynomial that ends with 1^6+2 = 1+2 = 3, which is the constant term of this polynomial.
The roots (a-1) through (f-1) multiply to 3 due to Vieta's formula mentioned.
Then,
243(a-1)(b-1)(c-1)(d-1)(e-1)(f-1)+1
= 243*3+1
= 730
is the final answer.
I used the CAS tool in GeoGebra to confirm it's correct.
here's the link to the workbook
https://www.geogebra.org/calculator/mrhj6tat
No matter which order you pick the roots, the final value will be the same each time. Your teacher is laying a trap when asking "find the largest possible value" to imply there could be several possible outcomes, which is not the case here.
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