Bringing A whole new level
For every positive integer g there is a unique factorial base expansion ( g 1 , g 2 , g 3 , … , g m ) , meaning that
g = 1 ! ⋅ g 1 + 2 ! ⋅ g 2 + 3 ! ⋅ g 3 + ⋯ + m ! ⋅ g m , where each g i is an integer, 0 ≤ g i ≤ i , and 0 < g m .
Given that ( g 1 , g 2 , g 3 , … , g j ) is the factorial base expansion of 1 6 ! − 3 2 ! + 4 8 ! − 6 4 ! + ⋯ + 1 9 6 8 ! − 1 9 8 4 ! + 2 0 0 0 ! , find the value of g 1 − g 2 + g 3 − g 4 + ⋯ + ( − 1 ) j + 1 g j .
The answer is 495.
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1 solution
That is not the number under consideration in the problem.
Write the number N = ( 2 0 0 0 ! − 1 9 8 4 ! ) + ( 1 9 6 8 ! − 1 9 5 2 ! ) + ⋯ + ( 4 8 ! − 3 2 ! ) + 1 6 ! . It is well known that n ! − 1 = k = 1 ∑ n − 1 k ⋅ k ! , so that n ! − m ! = k = 1 ∑ n − 1 k ⋅ k ! − k = 1 ∑ m − 1 k ⋅ k ! = k = m ∑ n − 1 k ⋅ k ! . Thus we have N = ( 1 9 9 9 ⋅ 1 9 9 9 ! + ⋯ + 1 9 8 4 ⋅ 1 9 8 4 ! ) + ( 1 9 6 7 ⋅ 1 9 6 7 ! + ⋯ + 1 9 5 2 ⋅ 1 9 5 2 ! ) + ⋯ + ( 4 7 ⋅ 4 7 ! + ⋯ 3 2 ⋅ 3 2 ! ) + 1 6 ! In other words, g i = ⎩ ⎪ ⎨ ⎪ ⎧ 1 i 0 i = 1 6 i ≡ 0 , … 1 5 mod 3 2 otherwise Now to do the final calculation. Note that 4 7 − 4 6 + 4 5 − 4 4 + − ⋯ − 3 2 = ( 4 7 − 4 6 ) + ⋯ + ( 3 3 − 3 2 ) = 8 ⋅ 1 = 8 ; There are 1 9 8 4 / 3 2 = 6 2 of these runs of 16 non-zero values, so that the final answer is 6 2 ⋅ 8 − 1 = 4 9 5 . (The final 1 is the term g 1 6 .)