4
$\begingroup$

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.

||improve this question
$\endgroup$
6
$\begingroup$

A solution with 5 twos:

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

Another solution with 5 twos:

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

||improve this answer
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

||improve this answer
$\endgroup$
  • $\begingroup$ I'm using base 10 ;) $\endgroup$ – Xandawesome Jun 21 '15 at 18:37
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ All numbers are base "10". :) $\endgroup$ – Jeremy Stein Jun 22 '15 at 17:11
  • 1
    $\begingroup$ In base 2, 2 doesn't exist. $\endgroup$ – Joel Rondeau Jun 26 '15 at 2:03
1
$\begingroup$

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 $22/.22$

||improve this answer
$\endgroup$
0
$\begingroup$

A solution with 6 twos and only subtraction and division:

$(222-22) \div 2$

||improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.