Linked to A four digit number using exact same 4 digits

A number with same repeated digits is called a Repdigit or Monodigit number https://en.wikipedia.org/wiki/Repdigit

Can you write an equation for the 7 digit Repdigit number 2222222 using only the digit 2 and its Repdigit numbers (EXCEPT the original number 2222222) and the following seven math operations:

+ - / * ! ^ and square-root

The seven math operations must be used exactly once.

Order of the math operations not important.

The RHS of the equation can only show Repdigit numbers of 2, including 2, but not 2222222.

Bonus: Can you create a general equation for the number 2222... that has "n" 2s and n>2?

  • $\begingroup$ Can I use parenthesis to change order of ops as desired? Also I'm assuming i can do something like 2sqrt(2) * 2!2, because two of the operations you gave only have one input and. $\endgroup$
    – Ankit
    Aug 18, 2021 at 21:24
  • $\begingroup$ OK so my solution does not involve paranthisis. And as for the two examples you gave @Ankit I think they involve Multiplication like 2!2 is really 2!*2 right? $\endgroup$
    – DrD
    Aug 18, 2021 at 21:40

2 Answers 2


If I've interpreted correctly then I think the following would work

$$\sqrt{2^2}+ 222222 *\left(\frac{22-2}{2!}\right) = 2222222$$

which generalises as follows

$$\sqrt{2^2}+ 22\ldots 2 *\left(\frac{22-2}{2!}\right) = 22\ldots 2$$

and if you are being strict about parentheses then perhaps we can change it to the following

$$2+ 222222 *\sqrt{\frac{22-2}{2!}}^2 = 2222222$$


My solution is on the same theme as @hexomino's solution but a bit simpler, I think

2222222 = 222222*22/2 - 222222 + sqrt2^2!

2222222 = 2444442 - 222220


2222...to"n" = 2222... to "n-1" *22/2 - 2222..to "n-1" + sqrt2^2! n >2

I have of course accepted @hexomino's answer


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