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

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.

## closed as too broad by McMagister, Fimpellizieri, hexomino, Alexis, Ben AaronsonMar 2 '16 at 20:46

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Can you use each digit more than once? – wythagoras Feb 25 '16 at 20:21

34/2.

• This is the answer to the updated question. – Ian MacDonald Feb 26 '16 at 15:16

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 

• you still have my upvote - most creative answer – Alex Feb 26 '16 at 15:23

Without concatenating and/or reusing digits:

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

• Good one !!!!!! – kavi temre Mar 2 '16 at 10:04

Using each number exactly twice

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

Exactly 17:

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

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

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

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

## 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.

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

• or take it one step further: 4 + 3 + 2 + 4 + 4 + 3 - 3 = 17 – Slepz Feb 25 '16 at 20:19
• @Slepz That was unnecessary. – ABcDexter Feb 25 '16 at 20:23
• @ABcDexter or take it one step further: 4 + 3 + 2 + 4 + 4 + 3 - 3 + 2 - 2 – Slepz Feb 25 '16 at 20:29
• @Slepz don't forget 4 - 4 + 4 + 3 - 3 + 3 + 2 - 2 + 2 + 4 - 4 + 4 + 4 - 4 + 4 + 3 - 3 + 2 - 2. – Nyk 232 Feb 26 '16 at 1:00

I try to make it simple !!

4/.2 - 3

And Another solution

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

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

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.

I guess this works:

$4^2 + 3 - 2$

• "using the numbers only once..." – ABcDexter Mar 1 '16 at 12:16
• oh sorry lol so the power is considered – Mekalikot Mar 1 '16 at 23:36