Make the number $100$ out of digits $1$, $2$, $3$, $4$, without repeats How can we make the number $100$, using only the following digits: $1$, $2$, $3$, $4$. You cannot repeat any of them.
 A: How about $(4!+1)(3!-2)=100$
I can't see anyway of doing this without factorials.
A: If factorial is allowed, then $(3!+4)^2$ could make the job (inspired by Nicholas).
A: If roots are allowed then: $\sqrt{(3^2+1)^4}$
A: Using only basic arithmetic:

$$ \frac{4\times 3}{.12} =100 $$

A: $$2\left[\left\lfloor\exp(4)\right\rfloor-(3+1)\right]$$
A: How about: $1^3 \cdot 4 = 100_2$
A: And solutions with logarithms (here natural logarithm) and ceil function.
$$\left\lceil \ln 2^{143}\right\rceil = 100
= \sqrt{\left\lceil \ln 2^{13}\right\rceil^4}$$

Edit: (I don't wan't to make other answer it's not that type of question) Or with binomial coefficient.
$$\binom{4+1}{3}^2 = 100$$
How to get it? Note that binomial coefficient equal to $10$ with small numbers are $\binom{5}{2}$ and $\binom{5}{3}$, but we have to have $2$ in power. So we just have to fin way to describe $\binom{5}{3}$ using $1,3,4$.

Edit2: I really didn't think about it! Just come to my mind. Here $\varphi$ is Euler's totient function.
$$ \varphi(312)+4 = 100$$
How to get it? Just note $1+2+3=6 = 3 \cdot 2$, so number made from this digits is divisible by 3. In very bad approximation $\varphi(n) \approx \frac{n}{3}$ and should be close to $100$, so we should check $312$. It could be too $214$ or $241$ (here very bad approximation for even $\varphi{(n)} = \frac{n}{2}$ and in fact $\varphi(214) = 106$). But it isn't.

Edit3: Ok, I was thinking a little bit now. It's last edit. Using summation and totient again.
$$ 12 +\sum_{k\mid34}\left(k + \varphi(k)\right) = 100 = 1 + 3 + \sum_{k \mid 42} k$$
A: What about $(1)(3!+4)^2$? Is factorial allowed? Is $2$ allowed as an exponent?
A: (4+3+2+1)^2. 
^2 is an exponent 
