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Let sigma(n) denote the sum of all positive divisors of n. Count the…
Question
Let sigma(n) denote the sum of all positive divisors of n. Count the number of integers n with 1 ≤ n ≤ 2000 for which sigma(n) is odd.
✓ Verified answer: 75checked by our engine — not a guess
Step-by-step solution
For n = 2^a · m with m odd, sigma(n) = sigma(2^a)·sigma(m).
sigma(2^a) = 2^(a+1)−1 is always odd.
For an odd prime power p^e, sigma(p^e) = 1+p+...+p^e is a sum of e+1 odd terms, which is odd iff e+1 is odd, i.e.
e is even.
Hence sigma(n) is odd iff every odd prime in n appears to an even power, i.e.
the odd part of n is a perfect square.
This happens iff n is a perfect square or twice a perfect square.
Count up to 2000: perfect squares ≤2000: floor(sqrt(2000))=44.
Twice a square ≤2000: 2k^2≤2000 ⇒ k^2≤1000 ⇒ k≤31, giving 31.
These two sets are disjoint (a square is never twice a square for positive integers).
Total = 44+31 = 75.
Final answer75
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