Boardwork

Probability Level 3

A student writes the six complex roots of the equation z 6 + 2 = 0 z^{6}+2=0 on the board. At every step, he randomly chooses two numbers a a and b b from the board, erases them, and replaces them with 3 a b − 3 a − 3 b + 4 3ab-3a-3b+4 . At the end of the fifth step, only one number is left. Find the largest possible value of this number.


The answer is 730.

Ryan Merino by Ryan Merino , 23, Philippines

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1 solution

Chris Lewis
Mar 3, 2021

Let f ( a , b ) = 3 a b − 3 a − 3 b + 4 f(a,b)=3ab-3a-3b+4 . First note we can rewrite this as f ( a , b ) = 3 ( a − 1 ) ( b − 1 ) + 1 f(a,b)=3(a-1)(b-1)+1

Now, f ( f ( a , b ) , c ) = 3 ( f ( a , b ) − 1 ) ( c − 1 ) + 1 = 9 ( a − 1 ) ( b − 1 ) ( c − 1 ) + 1 f(f(a,b),c)=3(f(a,b)-1)(c-1)+1=9(a-1)(b-1)(c-1)+1

It's easy to continue this process and see that whatever order we erase the numbers from the board, the last one will be X = 3 5 ( z 1 − 1 ) ( z 2 − 1 ) ( z 3 − 1 ) ( z 4 − 1 ) ( z 5 − 1 ) ( z 6 − 1 ) + 1 X=3^5 \left(z_1-1\right)\left(z_2-1\right)\left(z_3-1\right)\left(z_4-1\right)\left(z_5-1\right)\left(z_6-1\right)+1 where the z i z_i are the roots of g ( z ) = z 6 + 2 = 0 g(z)=z^6+2=0 . But the product ( z 1 − 1 ) ( z 2 − 1 ) ( z 3 − 1 ) ( z 4 − 1 ) ( z 5 − 1 ) ( z 6 − 1 ) = g ( 1 ) = 1 6 + 2 = 3 \left(z_1-1\right)\left(z_2-1\right)\left(z_3-1\right)\left(z_4-1\right)\left(z_5-1\right)\left(z_6-1\right)=g(1)=1^6+2=3

so that X = 3 6 + 1 = 730 X=3^6+1=\boxed{730} .

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