Question

Prime factorization of 1440: 1440 = 2⁵ × 3² × 5¹

Categorize all divisors of 1440 based on which prime factors they contain. Two divisors are coprime if and only if they share no common prime factors.

Organize the divisors into disjoint sets:

• S∅ = {1} (no prime factors)
• S₂ = {2, 4, 8, 16, 32} (only powers of 2)
• S₃ = {3, 9} (only powers of 3)
• S₅ = {5} (only powers of 5)
• S₂₃ = {6, 12, 18, 24, 36, 48, 72, 96, 144, 288} (powers of 2 and 3)
• S₂₅ = {10, 20, 40, 80, 160} (powers of 2 and 5)
• S₃₅ = {15, 45} (powers of 3 and 5)
• S₂₃₅ = {30, 60, 90, 120, 180, 240, 360, 480, 720, 1440} (all three primes)

The sizes of these sets are:

• |S∅| = 1
• |S₂| = 5
• |S₃| = 2
• |S₅| = 1
• |S₂₃| = 10
• |S₂₅| = 5
• |S₃₅| = 2
• |S₂₃₅| = 10

Two divisors are coprime if they come from sets that share no common prime factors. The valid combinations are:

1. S∅ with all other sets: The number 1 is coprime with every other divisor1 × (5 + 2 + 1 + 10 + 5 + 2 + 10) = 1 × 35 = 35 pairs

S₂ with sets containing no factor of 2: S₃, S₅, S₃₅

• 5 × (2 + 1 + 2) = 5 × 5 = 25 pairs

S₃ with sets containing no factor of 3: S₅, S₂₅

• 2 × (1 + 5) = 2 × 6 = 12 pairs

S₅ with sets containing no factor of 5: S₂₃

• 1 × 10 = 10 pairs

Total number of coprime pairs: 35 + 25 + 12 + 10 = 82

The formula of finding number of co-primes does not work here. *Just consider: *1 is coprime with ALL other 35 divisors. That's already 35 pairs, but the formula gives only 4 total!

4?

'4' is the answer of the question you stated above or anything else is asked in the real question of who's solution you have posted

6 should be the answer, no?

6?