Part one:

Compose number 6 with each digit using the digit only three times (Ex: 6=2+2+2) Allowed basic operations (+,-,*,/), sqrt and factorial (no others).

Part two:

Get number 13 using only number 2 three times. Allowed basic and not-basic operations.

(In reply to contribution to different ways by armando)

τ(√(₂2)) + H(2) = 13
τ(₂2) - 2$ = 13           

_nn = n^^^n : n pentated to n
     (₂2) = 2¹⁶ = 65536
√(n) : square-root of n
    √(65536) = 256
H(n) : hyperfactorial of n
    H(2) = 4
n$ : superfactorial of n           [as defined by Clifford Pickover]
    2$ = 4
τ(n) = tau(n) = d(n) : the number of divisors of n
    τ(256) = 9
    τ(65536) = 17

Posted by Dej Mar on 2018-03-18 09:39:23