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How many five-digit natural numbers have a digit sum equal to 43?
- 5
- 10
- 15
- 20
Correct answer: 15
Solution
Let d_i = 9 - e_i (i=1..5). Sum of d_i = 45 - sum(e_i) = 43, so sum(e_i) = 2. Each e_i in {0,...,9}. Since sum = 2, all e_i <= 2 automatically satisfying e_i <= 9. Also d₁ >= 1 means e₁ <= 8, automatically satisfied. So we count non-negative integer solutions to e₁+e₂+e₃+e₄+e₅ = 2 without restriction: C(2+4,4) = C(6,4) = 15.
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