Find the largest prime factor of 2 0 1 7 ! + 2 0 1 8 ! + 2 0 1 9 ! .
Notation:
!
is the
factorial
notation. For example,
8
!
=
1
×
2
×
3
×
⋯
×
8
.
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The above expression can be factorized as 2 0 1 7 ! ( 1 + 2 0 1 8 + 2 0 1 8 × 2 0 1 9 ) = 2 0 1 7 ! × 2 × 2 0 1 9 . Now The 1 + 2 0 1 8 + 2 0 1 8 × 2 0 1 9 = 4 0 7 6 3 6 1 Now 4 0 7 6 6 3 6 1 = 3 2 × 6 7 3 2 and 2 0 1 7 > 6 7 3
And 2 0 1 7 is a prime ∴ Answer comes out to be 2 0 1 7
The expression should be factored as 2 0 1 7 ! ( 1 + 2 0 1 8 + 2 0 1 8 × 2 0 1 9 ) =
2 0 1 7 ! ( 1 + 2 0 1 8 × 2 0 2 0 ) = 2 0 1 7 ! × 2 0 1 9 2 .
The answer is nevertheless still 2 0 1 7 as 2 0 1 9 is not a prime.
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Sir can u please post your solution. My method is tedious and time consuming. thank you
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2 0 1 7 ! + 2 0 1 8 ! + 2 0 1 9 ! = 2 0 1 7 ! ( 1 + 2 0 1 8 + 2 0 1 8 × 2 0 1 9 ) = 2 0 1 7 ! ( 2 0 1 9 + 2 0 1 8 × 2 0 1 9 ) = 2 0 1 7 ! ( 2 0 1 9 ) ( 1 + 2 0 1 8 ) = 2 0 1 7 ! ( 2 0 1 9 2 )
Since 2019 is divisible by 3 (its sum of digits is divisible by 3), it is not a prime. But 2 0 1 7 is a prime and it is the largest prime factor.