When Is the Sum of All Divisors an Odd Number? Olympiad Problem Explained
Hello everyone! Hope you’re enjoying the last days of summer. New posts in the channel have been a bit sparse lately.
But!
Changes are coming — more on that in September. The main thing is that activity here will pick back up soon: more mathematical beauty, math news, and food for thought. It’s useful to think every once in a while, especially when you have something to think with.
So stay tuned! In the meantime, here’s a problem to chew on — share your thoughts in the comments.
and this is an odd number. What is the smallest number N > 2025 for which the sum of all its divisors is also odd?
Answer: 2048
Why the Sum of Divisors Is Odd: The Key Theorem
To solve this, you need to know which numbers have an odd divisor sum in the first place.
The key fact from number theory: σ(n) is odd if and only if n is a perfect square or n is twice a perfect square.
Here is the idea behind this. Write n = 2^a × m, where m is odd. Then σ(n) = σ(2^a) × σ(m). The factor σ(2^a) = 1 + 2 + 4 + ⋯ + 2^a = 2^(a+1) − 1 is always odd. So the parity of σ(n) depends entirely on σ(m).
For odd m, σ(m) is odd if and only if m is a perfect square — because for each odd prime power p^k in the factorization of m, the sum 1 + p + p² + ⋯ + p^k has k + 1 terms (each odd), so it is odd precisely when k is even.
Putting it together: σ(n) is odd when n = 2^a × k² for some odd integer k. If a is even, this is a perfect square; if a is odd, this equals 2 × (perfect square).
Finding the Smallest N > 2025 With Odd Divisor Sum
Now we search for candidates just above 2025.
Perfect squares after 2025 = 45²: the next one is 46² = 2116.
Twice a perfect square after 2025: we need 2 × m² > 2025, so m² > 1012.5, meaning m ≥ 32. This gives 2 × 32² = 2 × 1024 = 2048.
Comparing the two candidates: 2048 < 2116, so N = 2048 is the answer.
Verification: Checking That σ(2048) Is Indeed Odd
2048 = 2^11, so its only divisors are powers of 2:
