Twos For Thought

Question

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.

score 6 · Accepted Answer · edited Jun 21 '15 at 10:48

6

A solution with 5 twos:

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

Another solution with 5 twos:

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

edited Jun 21 '15 at 10:48

Rand al'Thor

116,845
28
322
627

answered Jun 21 '15 at 04:36

pacoverflow

15,204
2
55
94

Good job! This seemed to be the lowest I could get too! – Xandawesome Jun 21 '15 at 06:20
@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. – Tom Carpenter Jun 22 '15 at 08:10

Rand al'Thor · Answer 2 · 2015-06-21T11:45:16.327

3

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.

edited Jun 21 '15 at 11:45

answered Jun 21 '15 at 10:57

Rand al'Thor

116,845
28
322
627

I'm using base 10 ;) – Xandawesome Jun 21 '15 at 18:37
1

@Xandawesome Ah, now you've edited to invalidate my answer! :'( ;-) – Rand al'Thor Jun 21 '15 at 18:38
Lol! +1 for creative thinking. – Xandawesome Jun 21 '15 at 18:40
1

All numbers are base "10". :) – Jeremy Stein Jun 22 '15 at 17:11
@JeremyStein ROFL! Very good point :-D – Rand al'Thor Jun 22 '15 at 17:12
1

In base 2, 2 doesn't exist. – Joel Rondeau Jun 26 '15 at 02:03

Quark · Answer 3 · 2015-06-26T03:14:32.823

1

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 $22/.22$

edited Jun 26 '15 at 03:14

answered Jun 26 '15 at 03:05

Quark

6,077
1
20
46

score 0 · Answer 4 · edited Jun 25 '15 at 23:55

0

A solution with 6 twos and only subtraction and division:

$(222-22) \div 2$

edited Jun 25 '15 at 23:55

Tryth

13,960
5
39
81

answered Jun 25 '15 at 22:00

Amy B

111
3

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

Twos For Thought

4 Answers4

Linked