Hello, digits!

The other day I came across a note from many years ago with some math problems, and one of the problems was to represent the number 6 using the same digit 4 times doing basic arithmetic operations. For example, 2 * 2 * 2 - 2 = 6 or 3 + 3 + 3 - 3 = 6. 6 is quite easy, and my note had the solutions using all digits, and then also 10 for some reason.

I looked at the note for a couple of seconds, and then a brilliant idea crossed my mind - why work if I can spend my time on an elementary school maths problem? So, I decided to solve this problem representing all numbers from 0 to 9. And I decided to try it with 3 digits as well.

The rule is that you use the same digit for 3 or 4 times and only basic arithmetic operations (+, -, *, /), square roots, factorials and parenthesis are allowed.

0 ✅ ✅

Representing 0 using the same digit is simple for any number of digits grater than 1.
For any integer n, n - n = 0 and n * 0 = 0 => n * (n - n) = 0.

# 3 digits:
✅ (n - n) * n = 0

# 4 digits:
✅ (n - n) * n * n = 0

1 ✅ ✅

1 is also simple.
For any integer n where n > 0, n / n = 1 and sqrt(n * n) = n. And for 0s we can use factorials, 0! = 1.

# 3 digits:
✅ 0! + 0 + 0 = 1
✅ sqrt(n * n) / n = 1

# 4 digits:
✅ 0! + 0 + 0 + 0 = 1
✅ (n * n) / (n * n) = 1

2 ✅ ✅

2 is still simple.
For any integer n where n > 0, (n + n) / n = 2. And for 0s we can again use factorials.

# 3 digits:
✅ 0! + 0! + 0 = 2
✅ (n + n) / n = 2

# 4 digits:
✅ 0! + 0! + 0 + 0 = 2
✅ (n + n) / sqrt(n * n) = 2

Patterns

After 2 there aren't really any simple formulas to get all the results, except for 3 with 4 digits. However there are some patterns we can use.

Same number, for any integer n, n + n - n = n. If we're solving for 5 we can use 5 + 5 - 5 = 5.
Previous and next numbers, for any integer n, where n > 2

(n - 1) + (n - 1) / (n - 1) = n
(n + 1) - (n + 1) / (n + 1) = n

If we're solving for 5, we can use 4 + 4 / 4 = 5 and 6 - 6 / 6 = 5.

Squares - 3 * 3 = 9 and sqrt(9) = 3, if there's a solution for one, it can be replicated for another one, and the same can be done for 2 and 4.

3 ❌ ✅

For any integer n where n > 0, 3 * n / n = 3 and we can use this for solving with 4 digits, but there's no formula for solving with 3 digits.

# 3 digits:
✅ 0! + 0! + 0! = 3
✅ 1 + 1 + 1 = 3
✅ 2 + 2 / 2 = 3
✅ 3 + 3 - 3 = 3
✅ 4 - 4 / 4 = 3
❌ 5 couldn't find a solution
❌ 6 couldn't find a solution
❌ 7 couldn't find a solution
❌ 8 couldn't find a solution
✅ sqrt(9) + sqrt(9) - sqrt(9) = 3

# 4 digits:
✅ 0! + 0! + 0! + 0 = 3
✅ (n + n + n) / n = 3

4 ❌ ❌

# 3 digits:
❌ 0 impossible
❌ 1 impossible
✅ sqrt(2 * 2) * 2 = 4
✅ 3 + 3 / 3 = 4
✅ 4 + 4 - 4 = 4
✅ 5 - 5 / 5 = 4
❌ 6 couldn't find a solution
❌ 7 couldn't find a solution
✅ sqrt(8 + sqrt(8 + 8)) = 4
✅ sqrt(9) + 9 / 9 = 4

# 4 digits:
✅ 0! + 0! + 0! + 0! = 4
✅ 1 + 1 + 1 + 1 = 4
✅ 2 + 2 + 2 - 2 = 4
✅ sqrt(3 * 3) + 3 / 3 = 4
✅ 4 + 4 * (4 - 4) = 4
✅ sqrt(5 * 5) - 5 / 5 = 4
✅ 6 - (6 + 6) / 6 = 4
❌ 7 couldn't find a solution
✅ sqrt(8 + 8) * 8 / 8 = 4
✅ 9 / sqrt(9) + 9 / 9 = 4

5 ❌ ✅

# 3 digits:
❌ 0 impossible
❌ 1 impossible
❌ 2 couldn't find a solution
✅ 3! - 3 / 3 = 5
✅ 4 + 4 / 4 = 5
✅ 5 + 5 - 5 = 5
✅ 6 - 6 / 6 = 5
❌ 7 couldn't find a solution
❌ 8 couldn't find a solution
✅ sqrt(9)! - 3 / 3 = 5

# 4 digits:
✅ (0! + 0! + 0!)! - 0! = 5
✅ (1 + 1 + 1)! - 1 = 5
✅ 2 + 2 + 2 / 2 = 5
✅ (3 + 3) / 3 + 3 = 5
✅ sqrt(4 + 4) + 4 / 4 = 5
✅ 5 * 5 / sqrt(5 * 5) = 5
✅ sqrt(6 * 6) - 6 / 6 = 5
✅ 7 - (7 + 7) / 7 = 5
✅ sqrt(8 + 8) + 8 / 8 = 5
✅ (9 + 9) / 9 + sqrt(9) = 5

6 ✅ ✅

# 3 digits:
✅ (0! + 0! + 0!)! = 6
✅ (1 + 1 + 1)! = 6
✅ 2 + 2 + 2 = 6
✅ 3 * 3 - 3 = 6
✅ 4 + 4 - sqrt(4) = 6
✅ 5 + 5 / 5 = 6
✅ 6 + 6 - 6 = 6
✅ 7 - 7 / 7 = 6
✅ 8 - sqrt(sqrt(8 + 8)) = 6
✅ 9 - 9 / sqrt(9) = 6

# 4 digits:
✅ (0! + 0! + 0! + 0)! = 6
✅ (1 + 1 + 1 * 1)! = 6
✅ 2 + 2 + 2 - 2 = 6
✅ 3 + 3 + 3 - 3 = 6
✅ (4 - 4 / 4) * sqrt(4) = 6
✅ sqrt(5 * 5) + 5 / 5 = 6
✅ 6 + 6 * (6 - 6) = 6
✅ sqrt(7 * 7) - 7 / 7 = 6
✅ 8 - (8 + 8) / 8 = 6
✅ 9 - sqrt(9) + 9 - 9 = 6

7 ❌ ❌

# 3 digits:
❌ 0 impossible
❌ 1 impossible
❌ 2 couldn't find a solution
✅ 3! + 3 / 3 = 7
❌ 4 couldn't find a solution
❌ 5 couldn't find a solution
✅ 6 + 6 / 6 = 7
✅ 7 + 7 - 7 = 7
✅ 8 - 8 / 8 = 7
✅ sqrt(9)! + 9 / 9 = 7

# 4 digits:
✅ (0! + 0! + 0!)! + 0! = 7
✅ (1 + 1 + 1)! + 1 = 7
❌ 2 couldn't find a solution
✅ 3 + 3 + 3 / 3 = 7
✅ 4 + 4 - 4 / 4 = 7
✅ 5 + (5 + 5) / 5 = 7
✅ sqrt(6 * 6) + 6 / 6 = 7
✅ 7 + 7 * (7 - 7) = 7
✅ sqrt(8 * 8) - 8 / 8 = 7
✅ 9 - (9 + 9) / 9 = 7

8 ❌ ❌

# 3 digits:
❌ 0 impossible
❌ 1 impossible
✅ 2 * 2 * 2 = 8
❌ 3 couldn't find a solution
✅ 4 + sqrt(4 * 4) = 8
❌ 5 couldn't find a solution
❌ 6 couldn't find a solution
✅ 7 + 7 / 7 = 8
✅ 8 + 8 - 8 = 8
✅ 9 - 9 / 9 = 8

# 4 digits:
❌ 0 impossible
❌ 1 impossible
✅ 2 + 2 + 2 + 2 = 8
✅ 3 * 3 - 3 / 3 = 8
✅ 4 + 4 + 4 - 4 = 8
❌ 5 couldn't find a solution
✅ 6 + (6 + 6) / 6 = 8
✅ sqrt(7 + 7) + 7 / 7 = 8
✅ 8 + 8 * (8 - 8) = 8
✅ sqrt(9) - 9 / 9 = 8

9 ❌ ❌

# 3 digits:
❌ 0 impossible
❌ 1 impossible
❌ 2 couldn't find a solution
✅ 3 + 3 + 3 = 9
❌ 4 couldn't find a solution
❌ 5 couldn't find a solution
❌ 6 couldn't find a solution
❌ 7 couldn't find a solution
✅ 8 + 8 / 8 = 9
✅ 9 + 9 - 9 = 9

# 4 digits:
❌ 0 impossible
❌ 1 impossible
❌ 2 couldn't find a solution
✅ 3 * 3 * (3 / 3) = 9
✅ 4 + 4 + 4 / 4 = 9
✅ 5 + 5 - 5 / 5 = 9
❌ 6 couldn't find a solution
✅ 7 + (7 + 7) / 7 = 9
✅ sqrt(8 + 8) + 8 / 8 = 9
✅ 9 + 9 * (9 - 9) = 9

That is it, but there are cases when it's impossible or I couldn't find a solution. In most cases they're impossible, I'll just need to prove them. I'll publish an update if I do that. I guess it will be in a couple of years, when I will accidentally stumble upon this article.