Inspired by Mohammad Rasel Parvej

Algebra Level 5

Let P ( x ) R [ x ] P(x) \in \mathbb{R}[x] be a Polynomial in x defined as follows: P ( x ) = x 2016 + x 2013 + x 2010 + x 2007 + . . . + x 6 + x 3 + 1 P(x) = x^{2016} + x^{2013} + x^{2010} + x^{2007} + ... + x^6 + x^3 +1 ; ( P : R R ) (P: \mathbb{R} \to \mathbb{R})

Let R ( x ) = C R(x)=C be the Polynomial that is obtained as remainder when P ( x ) P(x) is divided by x 3 3 x^3 -3

C C is a Constant Natural Number. Find the number of 1 1 's in Base-3(ternary base) in the representation of C C .


Inspiration

Bonus : How many 1's would have C C if P : Z 3 Z 3 P : \mathbb{Z}_3 \rightarrow \mathbb{Z}_3 ?


The answer is 673.

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1 solution

Otto Bretscher
May 10, 2016

We have P ( x ) 1 + 3 + 3 2 + . . . + 3 2016 / 3 m o d ( x 3 3 ) P(x)\equiv 1+3+3^2+...+3^{2016/3} \mod{(x^3-3)} , which is 673 \boxed{673} 1s when written in base 3. Note that x 3 3 x^3\equiv 3 so x 3 k 3 k x^{3k}\equiv 3^k .

Exactly,(+1) and the bonus?haha

Guillermo Templado - 5 years, 1 month ago

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We would be left with just one 1

Otto Bretscher - 5 years, 1 month ago

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Yes,that is all, folks... Thank you very much,Otto!

Guillermo Templado - 5 years, 1 month ago

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@Guillermo Templado What does Z 3 \mathbb{Z}_3 mean?

A Former Brilliant Member - 5 years, 1 month ago

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@A Former Brilliant Member Z 3 \mathbb{Z}_3 is a field, it's Z 3 Z \frac{\mathbb{Z}}{3\mathbb{Z}} . It's the set " { 0 , 1 , 2 } \{0,1,2\} " formed for the remainders of each integer modulus 3

Guillermo Templado - 5 years, 1 month ago

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@Guillermo Templado Ok. Thanks! So, in genral, Z n = { 0 , 1 , , n 1 } \mathbb{Z}_n=\{0,1, \cdots ,n-1\} right?

A Former Brilliant Member - 5 years, 1 month ago

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@A Former Brilliant Member Yes, exactly.

Guillermo Templado - 5 years, 1 month ago

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