Problems & Puzzles: Puzzles

Puzzle 906. 2018 Emmanuel Vantieghem sent the following nice equations equal to 2018  1 + 2 + 3 + 4*(5 - 6 + 7*8*9) = 2*(3 + 5 + 7*11*13) Q. Send other interesting ones for 2018?

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Contributions came from Flavio Torasso, Felipe G. Izquierdo and Emmanuel Vantieghem

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Flavio wrote on Dec 30, 2017:

2018 = (-2+3)*(-5+7)^11-13-17 =  2*(3!*5+7*11*13)-17+19-23-29-31+37

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Felipe wrote on Jan 4, 2018:

2018 =

2*3*5*7*11-13*17-19-23-29
2-3-5*7+11*13*17-19*23+29+31

2+3*5*7*11-13+17-19-23+29*31

31*29-23-19+17-13+11*7*5*3+2

31*29-23+19-17-13+11*7*5*3-2

31+29*23+19*17+13*11*7-5+3-2

31+29-23*19+17*13*11-7*5-3+2

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Emmanuel wrote  on Jan 4, 2018:

2018 = 3*674 - 9 + 5  (pandigital expression)

2018 = -2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23*29 + 31*37 + 41 + 43 + 47

2018 = 3^7 - 13^2  (four primes in use, not consecutive, not ordered)

2018 = 13^2 + 43^2  (unique sum of squares)
2018 = (9*8*7-6+5)*4+3+2+1

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Pierandrea Formusa wrote on Dec 9, 2018:

Below my solution for problem 906:
 

2018=PRIME(4)^LUCAS(4)-FIBONACCI(4)FIBONACCI(4+2)FIBONACCI(4)
 

where for Lucas and Fibonacci sequence LUCAS(1) =2 and FIBONACCI(1)=FIBONACCI(2)=1, PRIME(4) means 4th prime number (7), the expression involving fibonacci numbers it is a concatenation not a product: so 2018=7^4-383.

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