Friday Puzzler -- two dozen

Here are three

Larry Barrett

8+8+8

3^3-3

4!+4-4

good start

Alex Y

And I don't know how many there are. But one of yours suggests another:

4!*4/4

Re: Friday Puzzler -- two dozen

Henry Higginbotham

5P5/5

Okay, I don't know how to make a subscript on this forum.

Re: Friday Puzzler -- two dozen

Alex Y

spell it out. I don't understand what your formula is.

Re: Friday Puzzler -- two dozen

Henry Higginbotham

Permutation of 5 items from a set of 5 = 120, divided by 5 = 24.

At least, that's what I had in mind. My math background falls far short of most of the regulars here. Feel free to flunk me -- wouldn't be the first time!

Another, maybe

Henry Higginbotham

Now I'm even more out of my comfort zone regarding symbols on this forum, so I've had to screen capture it from CAD. You may have to look closely.

Re: Friday Puzzler -- two dozen

Alex Y

Very good. Should have thought of that interpretation.

Excellent!

Alex Y

You guys are coming up with ones I have not seen before.

Looking for another

Alex Y

that uses only the basic four arithmetic operators (and not all of them).

Hint: All of the solutions offered so far conform with a condition that is NOT a requirement of this puzzle. The solution I am thinking of violates the requirement you might have incorrectly inferred.

Re: Looking for another

Henry Higginbotham

Except for these variations on already stated themes, I'm ready to throw in the towel.

3! x 3 + 3!

4! x 4 / 4

8 + 8 - (-8)

-(-8 - 8 - 8 )

. . . etc.

Unless you mean we're freed from assuming the 24 is Base 10. If Base 7, then:

6+6+6

Nice solution

Alex Y

And you are right, I did not specify base 10.

There is another solution in base 10.

D'oh!

Henry Higginbotham

22+2 ?

That's the one I was looking for!

Alex Y

"three instances of the same digit" is a slightly less restrictive requirement than "three instances of the same one-digit number"

Good going Henry!