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Five Zeroes And ! Use To Make 120

Discussion in 'Online Games & Puzzles' started by Thyagarajan, Aug 5, 2020.

  1. Thyagarajan

    Thyagarajan IL Hall of Fame

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    :hello:USE FIVE ZEROES AND TRIGNO AND MATHS NOTATIONS TO MAKE 120
    I really love this puzzle. This is nothing but just requires your presence of mind and some tricky use of mathematical and trigonometric operators.

    1.) Make 120 with 5 zeros :-

    TRY AND SOLUTIONS SET DEEP BELOW


























    Sol 1 :- (0! + 0! + 0! + 0! + 0!)! = 120

    Here '!' symbol is used for factorial.
    And 0! = 1
    = (0! + 0! + 0! + 0! + 0!)!
    = (1 + 1 + 1 + 1 + 1)!
    = 5!
    = 120

    Sol 2 :- (cos (0) + cos (0) + cos (0) + cos (0) + cos (0))! = 120

    = (cos (0) + cos (0) + cos (0) + cos (0) + cos (0))!
    = (1 + 1 + 1 + 1 + 1)!
    = 5!
    = 120

    Similarly you can replace cos (0) with cot (0).

    Sol 3 :-(0!0! * 0!0!) - 0! = 120

    = (0!0! * 0!0!) - 0!
    = (11 * 11) - 1
    = 121 - 1
    = 120

    Sol 4 :- ((0! + 0!)^(0! + 0!) + 0!)! = 120

    = ((0! + 0!)^(0! + 0!) + 0!)!
    = ((1 + 1)^(1 + 1) + 1)!
    = (2^2 + 1)!
    = (4 + 1)!
    = 5!
    = 120

    Similarly you can do it in many ways. Seems quite easy. Isn't it. Lets make it a bit difficult.

    2.) Make 120 with 4 zeros :-

    ((0! + 0! + 0!)! - 0!)! = 120

    = ((0! + 0! + 0!)! - 0!)!
    = ((1 + 1 + 1)! - 1)!
    = (3! - 1)!
    = (6 - 1)!
    = 5!
    = 120

    3.) Make 120 with 3 zeros :-

    Γ((0! + 0! + 0!)!) = 120

    Here 'Γ' is called "Gamma Function".
    Γ (n) = (n - 1) ! , where n = 0, 1, 2, 3 , .......

    = Γ((0! + 0! + 0!)!)
    = Γ((1 + 1 + 1)!)
    = Γ(3!)
    = Γ(6)
    = 5!
    = 120

    4.) Make 120 with 1 zero :-

    Sol 1 :- sec ( tan-1 (…… sec ( tan-1 (0!)) ……)), with sec ( tan-1 ….) taken 14399 times = sqrt (14400) = 120

    Consider a right-angled triangle with sides (1,1,√2). Thus,
    √2 = sec ( tan-1( 0! ) )
    Now, consider a another right-angled triangle with sides (1,√2, √3). Here,
    √3 = sec ( tan-1 (sec ( tan-1( 0! ) ) ) )
    Extrapolating this idea futher, for any number x, we can represent √x using one 0 as,
    √x = sec ( tan-1 (…… sec ( tan-1 (0!)) ……)), where sec ( tan-1 ….) is taken x-1 times

    Sol 2 :- (((((0!)++)++)++)++)! = 120

    If you are known to programming then it'll be easy to understand.
    In C/C++ programming '++' operator stands for increment of 1.

    = (((((0!)++)++)++)++)!
    = (((((1)++)++)++)++)!
    = (((2++)++)++)!
    = ((3++)++)!
    = (4++)!
    = 5!
    = 120
     
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