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Let D be the set of all positive divisors of 2025. Compute the sum of…
Question
Let D be the set of all positive divisors of 2025. Compute the sum of those divisors d in D that satisfy d ≡ 1 (mod 4).
✓ Verified answer: 2821checked by our engine — not a guess
Step-by-step solution
2025 = 3^4 · 5^2, so its divisors are 3^a·5^b with 0≤a≤4, 0≤b≤2.
All such divisors are odd.
Modulo 4: 3≡3 and 5≡1, so 3^a·5^b ≡ 3^a (mod 4), which is 1 when a is even and 3 when a is odd.
We want a even: a ∈ {0,2,4}.
Sum over these of 3^a is 1+9+81 = 91; sum over b of 5^b is 1+5+25 = 31.
The total = 91·31 = 2821.
Final answer2821
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