Make the number 17 using only 2, 3, and 4 [closed]

Question

Is it possible to make the number 17 using only 2, 3, and 4?

Using the numbers only once and any operation, including factorials and other more advanced math.

And it would have to be exactly 17.

Answer 1

How about the following:

34/2.

Answer 2

Edit: This answer doesn't fit within the updated requirements.

Here you go:

      22     33333333
    2222           44
       2          44
       2          4
       2         44
       2         4 
       2         4 
   222222222     4 

Answer 3

Without concatenating and/or reusing digits:

$2^4 + \lfloor \sqrt{3} \rfloor$

Answer 4

Using each number exactly twice

$2^4 + 2^{4*(3-3)}$

Answer 5

Exactly 17:

$\lceil 4^2 - \cos(3) \rceil$

or slightly less (with an error of less than 0.06%):

$ 4^2 - \cos (3) \approx 16.98999 $

Answer 6

$$ \pi\left(\Gamma\left(\frac{\mathrm{archav}(S(\theta(\mathrm{sinc}(4))))}{\sqrt{\zeta(2)}}+\Im\left\{\log(\mathrm{erf}(-3))\right\}\right)\right)=17 $$

• $\pi$ is the prime-counting function
• $\Gamma$ is the gamma function
• $\mathrm{archav}$ is the inverse haversine
• $S$ is the successor function
• $\theta$ is the Heaviside step function
• $\mathrm{sinc}$ is the sinc function
• $\zeta$ is the Riemann zeta function
• $\Im$ is the imaginary part of a complex number
• $\mathrm{erf}$ is the error function

How it Works

When I saw:

any operation, including factorials and other more advanced math.

I knew I had to do something crazy using a few of my favorite functions. I started off with the prime-counting function. Since it always outputs a whole number, even for noninteger inputs, it gives me a bit of leeway. We just need to get in between $p_{17}=59$ and $p_{18}=61$, which conveniently bracket $60$, which seems like an easy target to reach.

I played around with $5!/2=\Gamma(6)/2=60$ for a while. Although it was easy to get the $5$ or $6$ I needed, I couldn't reach them using all the functions I wanted to include. Since I needed a divisor of $2$ (or $\sqrt{4}$) this left me with at most two terms, which was not enough.

Then I discovered that $\Gamma(\sqrt{6}+\pi)\approx 60.6657\ldots$. This is good for two reasons: first, it frees up the divisor outside the gamma function. Second, $\zeta(2)=\pi^2/6$, which gives me a chance to include the zeta function. At this point the expression in my head looks something like this:

$$ \pi\left(\Gamma\left(\frac{\pi}{\sqrt{\zeta(2)}}+\pi\right)\right) $$

(Note that $\pi\approx 3.1415\ldots$ is not the same as $\pi(\cdot)$ the prime-counting function.) I now just needed two ways to get $\pi$.

My idea for the first $\pi$ was to use obscure trigonometric functions. Fortunately I can use their inverses to get the angle $\pi$ without using the notation $f^{-1}$ due to the $\mathrm{arc}\cdots$ naming convention. I decided on the haversine, since $\mathrm{hav}(\pi)=1$ and $1$ is an easy number to reach. I wanted to get to the $1$ using the successor function, like $S(0)=1$. Now how do I get a zero? I used the unit step function with a negative argument, in this case $\mathrm{sinc}(-4)\approx-0.1892\ldots$.

For the second $\pi$ I wanted to play around with the logarithm and complex numbers; since $\log(z)=\log(|z|)+i \arg(z)$, any negative number gives an imaginary part of $\pi i$. For bonus points I wanted to approximate $\log(-1)=\pi i$. To get close to $-1$ I used the error function, which very quickly approaches $\pm 1$ for large inputs. In fact, $\mathrm{erf}(-3)\approx -0.999978\ldots$, and $\log(\mathrm{erf}(-3))\approx-0.000022\ldots+\pi i$. Of course, the real part doesn't matter since it gets chopped off anyway.

Answer 7

Easily: $$4 + 3 + 2 + 4 + 4 = 17$$

Answer 8

I try to make it simple !!

4/.2 - 3

And Another solution

(ln(e)/.2) + (4*3)

Answer 9

$2^{4}+\prod_{n=3}^{\infty}\frac{n}{n}$

Answer 10

Without using concatenation or rounding of any kind:

$$(2 \times 3)? - 4=17$$

Where

'?' is the "termial function" which is like factorial, but with addition instead of multiplication. Thus, $(2 \times 3)? = 6? = 1+2+3+4+5+6=21$. Take away 4 to get 17.

Answer 11

I guess this works:

$4^2 + 3 - 2$