Problem 10: Factoring Factorials

Number Theory Level 2

19 ! + 20 ! + 21 ! 19 ! = ( ( x − 3 ) ! − 3 ) 2 \large \frac{19!+20!+21!}{19!} = ((x-3)!-3)^2

Find x x .

Notation: ! ! denotes the factorial notation . For example: 8 ! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 8! = 1\times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 .


The answer is 7.

Ashwin R by Ashwin R , 16, USA

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1 solution

Chew-Seong Cheong
Apr 25, 2018

19 ! + 20 ! + 21 ! 19 ! = 19 ! ( 1 + 20 + 20 × 21 ) 19 ! = 1 + 20 + 20 × 21 = 21 ( 1 + 20 ) = 2 1 2 = ( 24 − 3 ) 2 = ( 4 ! − 3 ) 2 = ( ( 7 − 3 ) ! − 3 ) 2 \begin{aligned} \frac {19!+20!+21!}{19!} & = \frac {19!(1+20+20\times 21)}{19!} \\ & = 1+20 + 20\times 21 = 21(1+20) = 21^2 \\ & = (24-3)^2 = (4!-3)^2 = ((7-3)!-3)^2 \end{aligned}

Therefore, x = 7 x = \boxed{7} .

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