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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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