Let
f ( n ) = n times ∬ ⋯ ∫ n times d x d x ⋯ d x = a 0 x n + a 1 c 1 x n − 1 + a 2 c 2 x n − 2 + ⋯ + a n c n
where a i and c i are constants left over from performing the integration. Let g ( n ) = i = 0 ∑ n a i where a i are the constants in f ( n ) . Find the last three digits of g ( 1 0 ) .
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Observe a n = n !
Rest is easy.
i = 0 ∑ 1 0 i !
(mod 1 0 0 0 ) gives 9 1 4
f ( 1 0 ) has 1 0 indefinite integrals in a row.
This means that a 0 = 1 0 ! , a 1 = 9 ! , and so on, until a 1 0 = 0 ! .
∑ i = 0 1 0 i ! ≡ 9 1 4 ( m o d 1 0 0 0 )
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This is not a level 5 question. It should be marked level 4. But anyway the problem is nice.