Number Theory

Find the largest prime factor of 2017 ! + 2018 ! + 2019 ! 2017! + 2018! + 2019! .


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2017.

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2 solutions

2017 ! + 2018 ! + 2019 ! = 2017 ! ( 1 + 2018 + 2018 × 2019 ) = 2017 ! ( 2019 + 2018 × 2019 ) = 2017 ! ( 2019 ) ( 1 + 2018 ) = 2017 ! ( 201 9 2 ) \begin{aligned} 2017!+2018!+2019! & = 2017!(1+2018+2018 \times 2019) \\ & = 2017!(2019+2018 \times 2019) \\ & = 2017! (2019) (1+2018) \\ & = 2017! (2019^2) \end{aligned}

Since 2019 is divisible by 3 (its sum of digits is divisible by 3), it is not a prime. But 2017 \boxed{2017} is a prime and it is the largest prime factor.

Achal Jain
Mar 8, 2017

The above expression can be factorized as 2017 ! ( 1 + 2018 + 2018 × 2019 ) = 2017 ! × 2 × 2019 2017!(1+2018+2018 \times 2019) = 2017! \times 2 \times 2019 . Now The 1 + 2018 + 2018 × 2019 = 4076361 1+ 2018+ 2018 \times 2019 = 4076361 Now 40766361 = 3 2 × 67 3 2 40766361= 3^{2} \times 673^{2} and 2017 > 673 2017 > 673

And 2017 2017 is a prime \therefore Answer comes out to be 2017 2017

The expression should be factored as 2017 ! ( 1 + 2018 + 2018 × 2019 ) = 2017!(1 + 2018 + 2018 \times 2019) =

2017 ! ( 1 + 2018 × 2020 ) = 2017 ! × 201 9 2 2017!(1 + 2018 \times 2020) = 2017! \times 2019^{2} .

The answer is nevertheless still 2017 2017 as 2019 2019 is not a prime.

Brian Charlesworth - 4 years, 3 months ago

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Sir can u please post your solution. My method is tedious and time consuming. thank you

Achal Jain - 4 years, 3 months ago

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