Let P ( x ) ∈ R [ x ] be a Polynomial in x defined as follows: P ( x ) = x 2 0 1 6 + x 2 0 1 3 + x 2 0 1 0 + x 2 0 0 7 + . . . + x 6 + x 3 + 1 ; ( P : R → R )
Let R ( x ) = C be the Polynomial that is obtained as remainder when P ( x ) is divided by x 3 − 3
C is a Constant Natural Number. Find the number of 1 's in Base-3(ternary base) in the representation of C .
Bonus : How many 1's would have C if P : Z 3 → Z 3 ?
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Exactly,(+1) and the bonus?haha
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We would be left with just one 1
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Yes,that is all, folks... Thank you very much,Otto!
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@Guillermo Templado – What does Z 3 mean?
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@A Former Brilliant Member – Z 3 is a field, it's 3 Z Z . It's the set " { 0 , 1 , 2 } " formed for the remainders of each integer modulus 3
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@Guillermo Templado – Ok. Thanks! So, in genral, Z n = { 0 , 1 , ⋯ , n − 1 } right?
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@A Former Brilliant Member – Yes, exactly.
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We have P ( x ) ≡ 1 + 3 + 3 2 + . . . + 3 2 0 1 6 / 3 m o d ( x 3 − 3 ) , which is 6 7 3 1s when written in base 3. Note that x 3 ≡ 3 so x 3 k ≡ 3 k .