Challenge question.

You might want to download a free math editor like this one

https://math-editor.software.informer.com/download/

to present your answers.

I'd be glad to hear of any recommendations for a better one.

Along the same lines, you might like this. I enjoy playing it daily. There is a new puzzle every day at midnight GMT.

https://nerdlegame.com/

Now you've done it. I already waste my time on enjoy wordle and quordle every day.

Are zeroes allowed? (Technically a whole but not natural number)?
Must three digits be used if just one or two suffice?

I presume the allowed symbols include radical, parentheses, exponentiation, round-up and/or round-down, absolute value, decimal point and factorial (along with the obvious + - x ÷).

"Are zeroes allowed?"

Interesting, yes why not. (leads to a solution without any Ds)

"Must three digits be used if just one or two suffice"

No: although this is not always appropriate observe that d = √(d.d), so the question should read no more than three Ds.

"I presume the allowed symbols include radical"

Yes square root is OK but you can only use cube root if for example if you can create the little 3 (³√) using other allowable techniques.

Yes to all the others (you missed decimal point recuring which is very useful).

I'm not familiar with the recurring decimal point.

So far, the best I can do is four 1s, one 4, and two each of everything else.

My bad: perhaps I should have called it decimal point d recurring; usually shown by putting a dot above the d.

For example one neat solution for six is

Thanks. That is new to me; I'm familiar with an overscore, but I didn't think of it in my question about permissible operations.

The character set here does not have round-up or round-down symbols, but ↑ and ↓ work nicely, as in (((√6)↓)^((√6)↓))!

You mean like

3^3 - 3= 24?

-√9+(√9)^(√9)

8+8+8

Where is the challenge in this? and to what '9 problems' do you refet?

If you do the easy ones first, of course the challenge is less.

This can be done with one 4, two 2s or 5s, and three each of the other digits except 1.

I haven't yet found anything using three 1s.

Two 5s? I'llbe interested to see that.

Three 1s is relatively easy.

5!÷5

"Make 24 using three d s where d is a digit from 1 to 9."

Really any answer should include at least one solution for all nine digits.

But don't let that stop anyone from posting any individual answers as they find them.

The December Challenge is in the Blog Section.

Merry Christmas to all!

Finally I have found a solution with just two 1s, which I leave for now as part of the challenge.

OK, I used to try to rule out some things which seemed to be "cheating", but, for the most part that just suppressed creative thinking.

But this round up/round down thing does just seem to give up too many easy answers. So although I still want to see all answers I think you should try to get as many answers as you can without using rounding.

Here are a few answers