Mad factorials!
⎝ ⎛ n = 1 ∑ 2 0 1 7 ! ( n 2 0 1 7 ! ! ) ⎠ ⎞ 2 0 1 7 ! !
Find the last two digits of the expression above .
Notations :
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! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 .
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! ! denotes the double factorial notation:
For an even number & n > 0 , n ! ! = n × ( n − 2 ) × ⋯ × 4 × 2 .
For an odd number & n > 0 , n ! ! = n × ( n − 2 ) × ⋯ × 3 × 1 .
The answer is 76.
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1 solution
Which version of this problem is better ?
I think there's a mistake in pattern (2). If k = 1 and n = 3 , then ( 2 k + 1 ) n ≡ 3 3 ≡ 3 ≡ 1 ( m o d 8 ) . I think (2) should say something like ( 2 n + 1 ) ( 2 k + 1 ) ≡ 2 n + 1 ( m o d 8 ) or ( m o d 4 ) . Using this pattern, I found a different version of (1): x ≡ 1 , 7 ( m o d 8 ) ⟹ x ! ! ≡ 2 5 ( m o d 1 0 0 ) and x ≡ 3 , 5 ( m o d 8 ) ⟹ x ! ! ≡ 7 5 ( m o d 1 0 0 ) . Then, by (1) and (2), n = 1 ∑ 2 0 1 7 ! n 2 0 1 7 ! ! ≡ 1 + 7 5 + 7 5 + 2 5 + 2 5 + 7 5 + 7 5 + 2 5 + ⋯ + 7 5 + 2 5 ≡ 7 6 ( m o d 1 0 0 ) .
I observed some patterns :
(1) : x ≡ 1 ( m o d 8 ) ⇒ x ! ! ≡ 2 5 ( m o d 1 0 0 ) for x ≥ 1 5
(2) : ( 2 k + 1 ) n ≡ 1 ( m o d 8 ) where k ≥ 0 and n ≥ 2
(3) : ( ( 2 x ) 2 0 1 7 ! ! ) ≡ 0 ( m o d 1 0 0 )
By (3) :
n = 1 ∑ 2 0 1 7 ! + 1 ( n 2 0 1 7 ! ! ) ≡ 1 2 0 1 7 ! ! + 3 2 0 1 7 ! ! + 5 2 0 1 7 ! ! + 7 2 0 1 7 ! ! + … ( m o d 1 0 0 )
By (1) & (2) :
n = 1 ∑ 2 0 1 7 ! + 1 ( n 2 0 1 7 ! ! ) ≡ 1 + 2 5 + ⋯ + 2 5 ≡ 1 + ( 4 k − 1 ) ( 2 5 ) ≡ 7 6 ( m o d 1 0 0 )
⎝ ⎛ n = 1 ∑ 2 0 1 7 ! + 1 ( n 2 0 1 7 ! ! ) ⎠ ⎞ 2 0 1 7 ! ! ≡ 7 6 ( m o d 1 0 0 )
Answer = 7 6
Which version of this problem is better ?