I'm sure most of you are familiar with the famous four fours puzzle- which constists of trying to find ways to express every integer using only four digit 4s and specified operators. David A Wheeler, among others, have created compilations of solutions that stretch into five digits : https://www.dwheeler.com/fourfours/.

A similar problem is the five fives puzzle, which is basically the same, except that one uses five digits 5 instead of four digits 4. This puzzle is far less popular and I couldn't find any large compilations of solutions online, so I decided to attempt to create one myself (at least up to 1000). I am not going to post all of my solutions, but if you are curious about a particular solution, just message me and I'll be happy to tell you the answer I have found. I will, however, detail here, the values under 1,000 that I have not found solutions for with particular operators. I am offering a small prize for anybody who discovers a solution for a few specific numbers! See below for details.

I was able to find solutions for integers from 0 to 171 using only the elementary operators (addition, subtraction, multiplication, and division), plus powers, square roots, and the factorial. I also allowed concatenation and decimalization of the 5s, (e.g .5, 55, and 555 are acceptable. However, you cannot use concatenation as an operator, so don't try to represent 255 as 5*5 || 5). Many numbers beyond 171 have solutions using only these operators, but I believe 172 is the first number that requires something else.

To fill in the next missing values, such as 172, 223, and 227, I added two new things: the overline (so .5 with overline = .5repeating. This is the only application of the overline in the puzzle as there are no other digits, but this building block of 5/9 is extremely useful), and arbitrary roots. This does not mean cube roots or 6 roots are okay- they aren't. This means that you can use a root provided that you can represent the number using some of the 5s. So, for example, a .5root of a number is the equivalent of squaring it.

I found solutions for all but a handful of integers up to 732 using these new rules. The missing solutions were: 347, 353, 367, 373, 411, 453, 454, 463, 467, 493, 526, 527, 532, 548, 641, 642, 652, 668, 678, 679, 686, 689, 694, 698 and 707. I also found many solutions for numbers above 732, but I won't list the missing values as they are numerous.

To fill in the missing values, I introduced a new operator: the gamma function. If you aren't familiar with the gamma function, it is, in short, gamma(n) = (n-1)! It might seem a bit obscure to be included over something like %, but I chose it mainly because David Wheeler's paper considered it preferable to the percent and other operators. Also, % can be a bit ambiguous, as some calculators say that 100 + 5% = 100.05 (taking 5% to simply equal .05), while others say that 100 + 5% = 105 (taking + 5% to mean "add 5% of 100).

Anyways, introducing the gamma function allowed me to find solutions for nearly every integer up to 1000, and every integer up to 800....except 467. This missing value sticks out like a sore thumb.

The other missing values are: 802, 803, 813, 814, 827, 857, 871, 877, 878, 881, 886, 892, 897, 898, 903, 906, 914, 918, 919, 921, 926, 929, 934, 938, 942, 943, 946, 974, 978, 982, and 993.

Contest:I have devoted a lot of time and effort to finding solutions for this puzzle. Thus, the missing values really annoy me and I would appreciate any help in finding solutions. I am offering $5 to anyone who finds a solution to one of the missing values (or $10 for 467).

Disclaimers:

-You must use the specified operators. I already know of solutions for the numbers using other operators, such as double factorial or binomial coefficient, so giving a solution like that will be nothing new.

-I am not really looking for solutions for the numbers in the first list since I already have solutions (with gamma), but it would still be cool if you found a gamma-less solution to them. However, if you find a gamma-less solution to 347 (the smallest number that requires gamma), I will give you $5. This also holds for 885, which I believe to be the only multiple of 5 under 1000 that requires gamma.

-If you find multiple solutions, the maximum prize is $10 (or $15 if one of the solutions you found is 347). I will not give $5 per solution.

Tips and tricks:

-Create building blocks and write them down. For example, (sqrt(5 / .5repeating)!)! is a way to write 720 using only two fives, and is used in many solutions for numbers in the 650 to 770 range.

-Create a spreadsheet or program to look for solutions. I did this for some of the harder ones, but the spreadsheet was limited as it only searched for solutions with building blocks that I was aware of.

-Use negative exponents. .5^-5 is a simple way to represent 32.

-Use square roots to divide powers by two. sqrt(n^m) = n^(m/2). So, you can raise a number to the 24th power and take the square root of it three times to get the number to the power 3.

Thank you for reading and good luck if you take a crack at the problem!