Consider a sequence whose first term is a given positive integer \( N > 1 \). Consider the prime factorization of \( N \). If \( N \) is a power of 2, the sequence consists of a single term: \( N \). Otherwise, the second term of the sequence is obtained by replacing the largest prime factor \( p \) of \( N \) with \( p + 1 \) in the prime factorization. If the new number is not a power of 2, we repeat the same procedure with it, remembering to factor it again into primes. If it is a power of 2, the numerical sequence ends. And so on. For example, if the first term of the sequence is \( N = 300 = 2^2 \cdot 3 \cdot 5^2 \), since its largest prime factor is \( p = 5 \), the second term is \( 2^2 \cdot 3 \cdot (5 + 1)^2 = 2^4 \cdot 3^3 \). Repeating the procedure, the largest prime factor of the second term is \( p = 3 \), so the third term is \( 2^4 \cdot (3 + 1)^3 = 2^{10} \). Since we obtained a power of 2, the sequence has 3 terms: \( 2^2 \cdot 3 \cdot 5^2 \), \( 2^4 \cdot 3^3 \), and \( 2^{10} \). a) How many terms does the sequence have if the first term is \( N = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \)? b) Show that if a prime factor \( p \) leaves a remainder of 1 when divided by 3, then \( \frac{p+1}{2} \) is an integer that also leaves a remainder of 1 when divided by 3. c) Present an initial term \( N \) less than 1,000,000 (one million) such that the sequence starting from \( N \) has exactly 11 terms.