# Given any 44 numbers, is there a way to make 2424?

I was playing the game 2424, and I saw that some numbers, such as 5,8,9,105,8,9,10, could not be multiplied, divided, subtracted, added, etc. by me to get 2424 no matter how hard I tried…

Question: So that got me thinking: Is there a way to tell if a set of 44 numbers can be manipulated to make 2424?

You can use any operation sign and any operation you would like (meaning you can use log\log and derivatives, but using those is probably going to be inefficient).

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Looks like a good problem. (+1)
– Jacky Chong
Oct 21 at 2:13

Just a note: your four numbers 5,8,9,5,8,9, and 1010 cannot make 2424, so you can stop trying 😉
– suomynonA
Oct 21 at 2:14

@JackyChong Yes, I know, it was just a side note
– suomynonA
Oct 21 at 2:15

Isn’t this just the knapsack problem in disguise?
– John Douma
Oct 21 at 2:20

@suomynonA Yeah, don’t worry. I gave up right after I posted this problem…
– Frank
Oct 21 at 2:23

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If we are allowed to use the Gamma function:

Γ(5)⋅(9−8)10=24\Gamma(5)\cdot(9-8)^{10} = 24

or if we can use the factorial and square root:

(8105√9)!\left(8^{\frac{10}{5\sqrt{9}}}\right)!

I can probably think up some more using other special-ish functions. The way I learned 2424 we could use addition, subtraction, multiplication, division, exponentiation, and square roots (and, of course, parentheses). I’m pretty sure there isn’t a way to solve it with only those rules, although I’ll keep trying.

Edit: We have

√(8√9)105=24\sqrt{\left(8\sqrt{9}\right)^{\frac{10}{5}}} = 24

which uses only the “expanded standard” rules. It is the “cleanest” in my opinion of the solutions here.

Hm… I never learned about Gamma function (I’m only a student) but the factorial and square root solution was interesting!
– Frank
Oct 21 at 2:28

@Frank I’m only a student too, but you can never be too studenty to learn interesting math concepts. Also, thanks! The factorial function is usually omitted in 2424, if only for the reason that it becomes the find-44 game, but since you said we can use any functions I assumed it was fair game.
– Carl Schildkraut
Oct 21 at 2:29

Γ(n)=(n−1)!\Gamma (n) = (n-1)! for n=1,2,3,4,…n = 1,2,3,4,\ldots
– Alejo
Oct 21 at 2:32

@Frank I just found a “cleaner” solution – see my edit.
– Carl Schildkraut
Oct 21 at 2:57