4
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How many twos do you need to get to the number $100$ exactly?

Example: $22+22+22+22+2^2+2^2+2^2$

This example method uses 14 twos and is obviously not ideal. Can you find a way that uses the least amount of twos? I've seen some things like this on PuzzlingSE and they've been pretty well accepted so I decided to make my own, as these are fun to do! The lowest number of twos will win!

Math You Can Use:

  • Addition
  • Subtraction
  • Multiplication
  • Division
  • Exponents
  • Square & Cube Roots

Other Rules For Clarification:

  • The only number you may use are twos; any other numbers must be made with a combination of twos in some way.

  • This problem uses base 10 and the answer should be base 10.

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6
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A solution with 5 twos:

$(2\times2\times2 + 2)^2$

Another solution with 5 twos:

$\big(\frac{22-2}{2}\big)^2$

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  • $\begingroup$ Good job! This seemed to be the lowest I could get too! $\endgroup$ – Xandawesome Jun 21 '15 at 6:20
  • $\begingroup$ @Xandawesome If factorial is allowed, then this also works for a 5 twos solution: $ ((2^2)!\times2+2)\times2 $ . Couldn't come up with anything lower. $\endgroup$ – Tom Carpenter Jun 22 '15 at 8:10
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You never specified what base you were using...

In base 2, $100$ means the number four, so the answer is two twos ($2+2,2\times2,2^2$, etc.)
In base 3, $100$ means the number nine, which can be got with four twos: $\big(2+\frac{2}{2}\big)^2$.
In base 4, $100$ means sixteen, so the answer is three twos: $2^{2^2}$.
In base 5, $100$ means twenty-five, which can be got with five twos: $\big(2+2+\frac{2}{2}\big)^2$.
In base 6, $100$ means thirty-six, which can be got with four twos: $(2+2+2)^2$.

And so on.

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  • $\begingroup$ I'm using base 10 ;) $\endgroup$ – Xandawesome Jun 21 '15 at 18:37
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    $\begingroup$ @Xandawesome Ah, now you've edited to invalidate my answer! :'( ;-) $\endgroup$ – Rand al'Thor Jun 21 '15 at 18:38
  • $\begingroup$ Lol! +1 for creative thinking. $\endgroup$ – Xandawesome Jun 21 '15 at 18:40
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    $\begingroup$ All numbers are base "10". :) $\endgroup$ – Jeremy Stein Jun 22 '15 at 17:11
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    $\begingroup$ In base 2, 2 doesn't exist. $\endgroup$ – Joel Rondeau Jun 26 '15 at 2:03
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Edit: Here's three twos:

${\left(2\over{.2}\right)}^2$

Well since it's already been bumped, here it is, using four twos:

$(2*2)\over{(.2*.2)}$ as well as $(22-2)/.2$ and $22/.22$

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0
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A solution with 6 twos and only subtraction and division:

$(222-22) \div 2$

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  • $\begingroup$ That makes 200, not 100. $\endgroup$ – pacoverflow Jun 25 '15 at 22:19
  • $\begingroup$ @pacoverflow thanks - I fixed it, but now it has an extra 2 $\endgroup$ – Amy B Jun 25 '15 at 22:27
  • $\begingroup$ Amy B, I just gave you an approval for that amended answer. $\endgroup$ – Olive Stemforn Jun 26 '15 at 13:59

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