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All four numbers must be used for each solution, but they may appear in any order. Permitted:

  • the 4 basic mathematical operations,
  • square root symbol (maximum twice per solution, and the 2 is implied)
  • combining initial numbers by concatenation (eg. 5 and 7 can make 57)
  • exponentiation
  • decimal points, including omitting any initial zero
  • repeating decimal bar
  • parentheses
  • negative number sign.

Not permitted: everything else, including factorials, logarithms, infinite sums, rounding (floor, ceiling), numerical bases other than 10, concatenation of calculation results.

I have created all one hundred but had to use single factorials for seven of them. The missing ones for which non-factorial solutions are needed are 67 (now found), 87, 89 (now found), 91, 92, 94, and 99.

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    $\begingroup$ Additionally, are you asking for help or posting this as a challenge? If you are asking for help it helps to give the answers for 65~100 on the whole. $\endgroup$
    – Number Basher
    Jun 12, 2022 at 12:29

4 Answers 4

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Here's an answer for 67:

$67 = \dfrac{(7+.5)}{\overline{.1}}-.5$

I'll edit in any others as I find them.

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89:

$$ \frac{.5^{-5}}{\sqrt{.\overline{1}}} - 7 $$ $$ = \frac{32}{\left(\frac{1}{3}\right)}-7 = 96-7 = 89$$

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  • $\begingroup$ Bass, this solution for 89 will replace my factorial-containing 5! x .75 - 1. $\endgroup$
    – Jay
    Jun 20, 2022 at 13:57
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    $\begingroup$ Hmm, 0.75 is valid? I thought this would be a concatenation $\endgroup$
    – klabuster_
    Jun 20, 2022 at 14:13
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Here is my answer for $89$

For $89$ (if this is allowed) $155_7$ - meaning $155 (\text{base} 7) = 1\times49+5\times7 + 5\times1 = 49+ 35+5 = 89$

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    $\begingroup$ If the complement function defined as $\sim a = -a-1$ where $ a\in\mathbb{Z}$ is allowed then, I have solutions for all of the missing ones $\endgroup$
    – Suzuna Minami
    Jun 12, 2022 at 17:50
  • $\begingroup$ no base convertion is allowed $\endgroup$
    – Rafe
    Jun 23, 2022 at 7:48
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2 partial solutions for 99:

$$ \frac{7+\frac{1}{.5 \times .5}}{.\bar{1}} = \frac{7 + 4}{\frac{1}{9}} = 99$$

$$ \frac{\frac{1}{.\bar{5}-.5} - 7}{.\bar{1}} = \frac{18-7}{\frac{1}{9}} = 99$$

Both of them use an extra 1 which I cannot eliminate, maybe someone else can.

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