# Create all of the integers from 1 to 100 using 1, 5, 5, and 7 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.

• 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. Jun 12 at 12:29

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

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

89:

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

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

Here is my answer for $$89$$

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

• 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 Jun 12 at 17:50
• no base convertion is allowed
– Rafe
Jun 23 at 7:48

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.