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Each of the 5 factors is irreducible polynomial, so the problem is equivalent with finding the number of divisors. Considering each of the 5 factors as divisors (a, b, c, d, e), we obtain 32 divisors, so there are 32 cyclic codes of block length 15. We also recall that when multiplying two of the factors, the operation is preserved (we also obtain a cyclic code from the product of two or more cyclic codes)....
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