Can You Solve The Three 3s Challenge?

Thanks to Yale from Hong Kong for suggesting this problem! This problem went viral after Carl Ho's video (in Chinese), and there are many fun ways to solve it. My challenge is: find two answers to 3 3 3 = 10 using only the symbols + ! ( ). There is a way you can do this!

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Sources

Can You Solve The 6s Challenge?

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Connect with me

[CarlHoHK]

Glad to see my question in this channel :D
and yes, the "subfactorial" should be the better way to solve =10 question.

[arjunck]

Arjun(3+3+3) = 10
Arjun is a function created by me which adds 1 to the result

[amit_bisht]

Easy!
(3+3+3)++ = 10

As notation used by programmers seems to be acceptable.

[SebastienPatriote]

3.3 repeating (use a bar on top of the second three) multiplied by 3.

3.3....×3=10

[Buttered.Coffee]

When you were showing alternate solutions for 0-9, I saw a decimal point being used for 7 and I realized an easy way to do 10 is:
(3! - 3) / .3 = 10

[abysslight]

I definitely think using a decimal point as an operator is definitely cheating. .3 is not an operator acting upon 3, it is simply a different number. You can make a similar case for 3% as well, it is simply a different number.

[nasekiller]

the only hard thing about the bonus is knowing that subfactorial exists, lol.

[sharifikiasmethodsinmathem]

The other eleven relationships are:
☆ncr(3!-sgn(3), 3)=10
3×3+sgn(3)=10
3!+3+[Ln3]=10
3×3+[arcsec3]=10
[Ln(3+3)^3!]=10
((The symbol [ ] is the correct bracket or component function.))
Also, the following six relations are based on the number of divisors of the numbers:
3!+d(3^3)=10
d(3^(3×3))=10
3^d(3)+sgn(3)=10
(3!-sgn(3))×d(3)=10
ncr(3!, 3)÷d(3)=10
npr(3!, d(3))÷3=10
In the above relations, d(n) is the number of divisors for natural number, n.
We begin by writing the number as a product of prime factors: n = p^a× q^b× r^c...
then the number of divisors, d(n) = (a+1)(b+1)(c+1)...
To prove this, we first consider numbers of the form,  n = p^a. The divisors are 1,  p,  p^2, ...,  pa; that is, d(p^a)=a+1.
According to the law, we must act in such a way that we only use the number 3, 3 times, and therefore we can not use decimal, non-3 and radical power, which is a kind of power.

[BubblewrapMe]

I don’t think casually changing 3 to 0.3 in one of your solutions is acceptable.

[MinusPi-pc]

The set of allowed operators really should've been defined before asking for solutions. The subfactorial thing for 10 makes it seems like it was a trick question. Everyone knows ! as the factorial, and hardly anyone knows about the subfactorial, so those who do might think you only mean to use that factorial. Hell, without a defined set of operations, I could define my own operation ? that just brings everything to 10: 3?+3-3=10. Done!

[TyTheRegularMan]

I was expecting double factorial, so I found this:

(3!)!! = 6!! = 6*4*2 = 48
((3!)!!)/3 = 16
((3!)!!)/3 - 3! = 16 - 6 = 10

[Spax]

No rules specifically mentioned about adding operations to the right side of the equal sign:
(( 3 - 3 )/ 3 )! = log(10)
Also allows for other neat stuff such as:
( 3 + 3 )!/ 3 ! = 5!

[davidhutagaol]

who else didn't know the term subfactorial exist until this day?

[jarlfenrir]

3*3+sgn(3) = 10
Simple and elegant. Since we can use arctan, using signum fucntion should also be allowed.

[Me-da-Ghost]

For "0", you could just do (3 - 3) * 3.

[lordrindfleisch]

The derivative solution for 10 was really cool

[Yerflua]

Old memes: on the Pentium, 3+3+3 = 10 for very large values of 3

[carlosfermin]

At the beginning of the video, the instructions say: “Make each equation true using mathematical operations”. So I just did:
3! / (3+3) = 1+0
Hahahahahaha

[ericintheclouds]

I see in a lot of the solutions for 3 3 3 = 10 that the square root was taken. This is the same as raising to the power of 1/2 which leads me to assume that I can raise a 3 to any power I want...like 0. this means I can say (3 x 3) + 3^0 = 10

[tolipydob]

If we follow the convention used for hexadecimal number representation but use base twenty seven then 3 ( 3 ( 3 ) ) = 10 or 3 * 3 * 3 = 10
In base nine, 3 + 3 + 3 = 10
If we are already shifting bits in previously displayed examples, we can also choose to not limit ourselves to 10(base two) numbering systems.
10 has a disguise for every day of the week.