My First Problem

Number Theory Level 3

Let

X = ∑ n = 1 1000 n ! ⋅ n X = \sum_{n=1}^{1000} n! \cdot n

What is the remainder when X X is divided by 1001?


The answer is 1000.

Antonio Hugo by Antonio Hugo , 22, Indonesia

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1 solution

Antonio Hugo
Jan 5, 2016

∑ n = 1 1000 n ! n \sum _{ n=1 }^{ 1000 }{ n!n }

Let n = N − 1 n=N-1

∑ N = 2 1001 ( N − 1 ) ! ( N − 1 ) \sum _{ N=2 }^{ 1001 }{ (N-1)!(N-1) }

= ∑ N = 2 1001 N ( N − 1 ) ! − ( N − 1 ) ! =\sum _{ N=2 }^{ 1001 }{ N(N-1)!-(N-1)! }

= ∑ N = 2 1001 N ! − ( N − 1 ) ! =\sum _{ N=2 }^{ 1001 }{ N!-(N-1)! }

= ( 2 ! − 1 ! ) + ( 3 ! − 2 ! ) + . . . . . . . . + ( 1001 ! − 1000 ! ) =(2!-1!)+(3!-2!)+........+(1001!-1000!)

= 1001 ! − 1 = 1001!-1

X = 1001 ! − 1 X=1001!-1

Thus,

X ( m o d 1001 ) = 1001 ! − 1 ( m o d 1001 ) ≡ − 1 ( m o d 1001 ) ≡ 1000 ( m o d 1001 ) X(mod\quad 1001)= 1001!-1(mod\quad 1001)\equiv -1(mod\quad 1001)\equiv \boxed{1000(mod\quad 1001)}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I just can watch how u solve this:")

Jessill Jess - 4 years, 7 months ago

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