Multiplying Large Polynomials

We define the functions p o l y R e p \mathrm{polyRep} and s t r R e p \mathrm{strRep} in the context for polynomials with coefficients from G F ( 2 ) GF(2) as follows:

s t r R e p ( i = 0 n 1 a i x i ) = a 0 a 1 a n 1 p o l y R e p ( a 0 a 1 a n 1 ) = i = 0 n 1 a i x i \mathrm{strRep}(\sum_{i=0}^{n-1}a_ix^i) = \overline{a_0a_1 \cdots a_{n-1}} \\ \mathrm{polyRep}(\overline{a_0a_1 \cdots a_{n-1}}) = \sum_{i=0}^{n-1}a_ix^i

Given two n n -bit strings A A and B B , and another n + 1 n+1 bit string P P , we define their P P product as

P r o d P ( A , B ) = s t r R e p ( c × p o l y R e p ( B ) ( m o d p o l y R e p ( P ) ) ) \mathrm{Prod}_P{(A, B)} = \mathrm{strRep}\left (c \times \mathrm{polyRep}(B) \pmod{\mathrm{polyRep}(P)}\right )

where the multiplication is considered as multiplication of polynomials over G F ( 2 ) GF(2) .

A natural definition of exponentiation is immediate:

E x p P ( A , k ) = s t r R e p ( p o l y R e p ( A k ) ( m o d p o l y R e p ( k ) ) ) \mathrm{Exp}_P(A,k) = \mathrm{strRep} \left ( \mathrm{polyRep}(A^k) \pmod{ \mathrm{polyRep}(k)} \right )

Compute E x p P ( A , k ) \mathrm{Exp}_P(A,k) for the given values and let r r be the result. Enter your answer as the number of 1s in r r .

1
2
3
P = 10000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000001
A = 0000100011010110011100111011001001001101000100100011111011110001010001000101111101101010000100110010
k = 2017


This problem is taken from a list of problems posted by RC Bose Center of Cryptography.


The answer is 56.

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