3
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  1. Complete the equality to make it true:

    1 1 1 1 1 = 945

  2. Using similar logic, solve for this:

    1 1 1 1 1 = 80

Info:

  • You should modify only the left side of =
  • The equation must remain an equality.
  • You can use any operator you want, as many parentheses as you like, combinations, decimals, powers and roots of any value etc.
  • There is at least 1 guaranteed solution for each

Be creative :)

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  • $\begingroup$ combining 1s is possible, such as 11+111? $\endgroup$ – Oray Jun 17 '17 at 13:09
  • $\begingroup$ yes, it is, if you'd like that $\endgroup$ – jack Jun 17 '17 at 14:35
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1. 945

$(1+(1+1+1)!)!!\div.\bar{1}=945$, where $!!$ denotes double factorial. [Calculation]

2. 80

$((1+1+1\times 1)!)!\times .\bar{1}=80$. [Calculation]

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  • $\begingroup$ you got the idea. Now try without using the dash (vinculum) :D $\endgroup$ – jack Jun 17 '17 at 14:42
2
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This was my answer:

1. 945

$((1+1+1+1)!! + 1)!! = 945$

2. 80

$((1+1+1)! + 1 + 1)!!! = 80$, where 8!!! = 8 x 5 x 2

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