2018 Factorials Madness

Number Theory Level 2

Determine the smallest positive integer n n such that the product of n n and 1 ! 2 ! 3 ! … 2016 ! 2017 ! 2018 ! 1009 ! \frac{1!2!3!\dots2016!2017!2018!}{1009!} will be a perfect square.


The answer is 2.

Valentio Iverson by Valentio Iverson , 17, Indonesia

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1 solution

X X
May 23, 2018

1 ! 2 ! 3 ! … 2016 ! 2017 ! 2018 ! 1009 ! \frac{1!2!3!\dots2016!2017!2018!}{1009!} = ( 1 ! ) 2 × 2 × ( 3 ! ) 2 × 4 × ( 5 ! ) 2 × 6 × ⋯ × ( 2017 ! ) 2 × 2018 1009 ! =\frac{(1!)^2\times2\times(3!)^2\times4\times(5!)^2\times6\times\cdots\times(2017!)^2\times2018}{1009!} = ( 1 ! 3 ! 5 ! 7 ! ⋯ 2017 ! ) 2 × 2 × 4 × 6 × ⋯ × 2018 1 × 2 × 3 × ⋯ × 1009 =(1!3!5!7!\cdots2017!)^2\times\frac{2\times4\times6\times\cdots\times2018}{1\times2\times3\times\cdots\times1009} = ( 1 ! 3 ! 5 ! 7 ! ⋯ 2017 ! ) 2 × 2 1009 = ( 1 ! 3 ! 5 ! 7 ! ⋯ 2017 ! × 2 504 ) 2 × 2 =(1!3!5!7!\cdots2017!)^2\times2^{1009}=(1!3!5!7!\cdots2017!\times2^{504})^2\times2

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