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How can we make the number $100$, using only the following digits: $1,2,3,4$. You cannot repeat any of them.

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    $\begingroup$ Do you know that there is a solution at all? $\endgroup$ – rewritten Dec 2 '14 at 22:12
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    $\begingroup$ Yes there is forsure an answer $\endgroup$ – Sarah Dec 2 '14 at 22:18
  • $\begingroup$ $((1+4)\times2)^3$ gives $1000$, if that helps. $\endgroup$ – Akiva Weinberger Dec 2 '14 at 22:26
  • $\begingroup$ $(1+23)\times4$ gives $96$… so close. (Am I allowed to write $23$?) $\endgroup$ – Akiva Weinberger Dec 2 '14 at 22:30
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    $\begingroup$ $3^4 + 21 = 102$ is as close as you're going to get, I think, unless you allow factorials or square roots or some other operation. $\endgroup$ – mjqxxxx Dec 2 '14 at 22:38
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Using only basic arithmetic:

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

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How about: $1^3 \cdot 4 = 100_2$

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  • $\begingroup$ I don't think it's valid. $100_{10} \neq 100_2$ But idea with other systems is nice. However I didn't find any way to use them. $\endgroup$ – Tacet Dec 2 '14 at 22:54
  • $\begingroup$ I think writing base 2 on L.H.S. would be better $\endgroup$ – sai kiran grandhi Dec 6 '14 at 7:02
  • $\begingroup$ @saikirangrandhi Och, yea. But have you any idea how? $\endgroup$ – Tacet Dec 7 '14 at 16:43
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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.

$$ 12 +\sum_{k\mid34}\left(k + \varphi(k)\right) = 100 = 1 + 3 + \sum_{k \mid 42} k$$

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    $\begingroup$ There's also ${(\sigma(23)+1)} \times 4 = 100$ :-) (using the divisor function) $\endgroup$ – Guri Harari Dec 2 '14 at 23:44
  • $\begingroup$ Nice, I was thinking about totient, but was too lazy to plug in numbers to see what would pop up. $\endgroup$ – Hao Ye Dec 2 '14 at 23:45
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    $\begingroup$ Could also use $p_{24+1}+3$, where $p_i$ is the $i$th prime number. $\endgroup$ – Hao Ye Dec 2 '14 at 23:53
  • $\begingroup$ @GuriHarari Nice, you can improve your answer, using edit. $\endgroup$ – Tacet Dec 2 '14 at 23:54
  • $\begingroup$ @HaoYe: I just thought isn't $\varphi(312) = 96 = 100-4$? I wasn't checking numbers. You can also improve your answer. $\endgroup$ – Tacet Dec 2 '14 at 23:55
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What about $(1)(3!+4)^2$? Is factorial allowed? Is $2$ allowed as an exponent?

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  • $\begingroup$ this gives 125... $\endgroup$ – rewritten Dec 2 '14 at 22:09
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    $\begingroup$ I am a little weak on arithmetic. $\endgroup$ – André Nicolas Dec 2 '14 at 22:19
  • $\begingroup$ I'm sure you aren't! (btw, 125 was the result of the original answer, this one gives in fact 100) $\endgroup$ – rewritten Dec 2 '14 at 22:20
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If factorial is allowed, then $(3!+4)^2$ could make the job (inspired by Nicholas).

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  • $\begingroup$ I'm only in 7th grade don't know what that means $\endgroup$ – Sarah Dec 2 '14 at 22:18
  • $\begingroup$ $3!$ means that you multiply $3\times 2\times 1$ (which is 6 then). $\endgroup$ – Peter Franek Dec 2 '14 at 22:19
  • $\begingroup$ Ohh okay! That's a great example but now sure if that's allowed do u know a way without it? $\endgroup$ – Sarah Dec 2 '14 at 22:22
  • $\begingroup$ I don't, unfortunately.. $\endgroup$ – Peter Franek Dec 2 '14 at 22:22
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    $\begingroup$ @Sarah I strongly believe, there is no way to do it with only +,-,*,/. $\endgroup$ – Tacet Dec 2 '14 at 22:39
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How about $(4!+1)(3!-2)=100$

I can't see anyway of doing this without factorials.

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If roots are allowed then: $\sqrt{(3^2+1)^4}$

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    $\begingroup$ No roots can u find away only using numbers exponents and any math sign (ex. 23+1-4= 20) $\endgroup$ – Sarah Dec 2 '14 at 22:29
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    $\begingroup$ i'm pretty sure that if you allow only +,-,x,/ and exponents than there's no solution. $\endgroup$ – Guri Harari Dec 2 '14 at 22:43
  • $\begingroup$ @GuriHarari There is a solution. See my answer. $\endgroup$ – jlars62 Dec 3 '14 at 17:35
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$$2\left[\left\lfloor\exp(4)\right\rfloor-(3+1)\right]$$

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(4+3+2+1)^2.

^2 is an exponent

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    $\begingroup$ Notice that this uses the digit $2$ twice. $\endgroup$ – user296602 Jan 3 '16 at 19:45

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