Question

Mathematics puzzle. Express numbers in terms of 2's with a maximum of four 2's.

Answer 1

Starting from one.

Answer 2

1 = 2 - 2 + 2 / 2.

Answer 3

2 = 2 / 2 + 2 / 2

Answer 4

3 = (2 + 2 + 2)/2

Answer 5

4 = 2 + 2 + 2 - 2.

Answer 6

And 5? :P

Answer 7

5 = 2 + 2 + 2 / 2. :)

Answer 8

6/2 is not 5 lol

Answer 9

Nope. It's 2 + 2 + 2 / 2 = 2 + 2 + 1 = 5. :))

Answer 10

Oh I see. Nice.

Answer 11

gets difficult as the numbers get bigger.

Answer 12

Yeah. :))

Answer 13

Spin-off Question: What is the largest integer that can be expressed using 4 2's and any mathematical operation?

Answer 14

There is a general formula to express any integer in terms of 2's with a maximum of 4 2's. I dunno it

Answer 15

Lets jump to 10!

Answer 16

10 = 2 x 2 x 2 + 2

Answer 17

@iambatman You came and left. Get back here! :P

Answer 18

11?

Answer 19

This one is hard. But, \[11=\frac{(2 + 2)! -2}{2}.\]

Answer 20

Ingenious!

Answer 21

Thanks. :))

Answer 22

\[12= 2(2^{2}+2)\]

Answer 23

\[\text{or }12=2 \cdot(2 + 2/2)!.\]So complicated LOL.
I can't think for 13.

Answer 24

Me too

Answer 25

@pgpilot326 said 13 = 22/2 + 2.

Answer 26

\[
6=\frac{(2+2)!}{2+2}
\]

Answer 27

\[2^{2^{2}}-2 = 14\]

Answer 28

Nice one @pgpilot326

Answer 29

also \[\frac{(2+2)!}{2}+2=14

Answer 30

oops \[\frac{(2+2)!}{2}+2=14\]

Answer 31

A medal for working for 13. :)

Answer 32

I figured out the one for 15 but it used 5 2's.

Answer 33

Answer to the spin-off question: \(2^{2^{22}}\). I don't know how it was obtained, but that's the answer I saw. :)

Answer 34

\[\sqrt{22^2}-2=20\]

Answer 35

Easy \[16=2^{3}\times2\]

Answer 36

An alternative way for 16 is welcome.

Answer 37

\[2^2\cdot 2^2 = 16\]

Answer 38

\[(2+2)^2=16\]

Answer 39

Even better using 3 two's.

Answer 40

\[2^{2^{2}}=16\]

Answer 41

\[15= (2+2)^{2}-2^{0}\]

Answer 42

so you can use numbers besides 2?

Answer 43

No. I broke the rule.

Answer 44

\[\left(\begin{matrix}2+2+2 \\ 2\end{matrix}\right)=15\]

Answer 45

17

Answer 46

22-2-2=18

Answer 47

\[(2\cdot 2+\frac{2}{2})!!=15\]

Answer 48

\[(2+\frac{2}{2})^2=9\]

Answer 49

Can you get 7?

Answer 50

\[(2+2)!!-\frac{2}{2}\]

Answer 51

also \[(2+2)!!+\frac{2}{2}=9\]

Answer 52

What is !!?

Answer 53

double factorial

Answer 54

\[3!! = 1\cdot 3\] etc.

\[4!!=2\cdot 4\] etc.

Answer 55

0!!=1

Answer 56

22/2 = 17 in hex

Answer 57

doesn't have to be base 10, right?

Answer 58

and then 22/2 + 2 = 19 in hex

Answer 59

The only rule I know is that you must use a maximum of 4 2's

Answer 60

winner, winner, chicken dinner!

Answer 61

i'm gonna use this in my classes... what a great puzzle!

Answer 62

22-2=20, 22-2/2=21, 22=22, 22+2/2=23, (2+2)!=24, (2+2)! + 2/2 = 25
22+2+2 = 26

Answer 63

Are you a maths teacher?

Answer 64

sometimes

Answer 65

2(22-2)=28 in base 7
(2+2)! + 2 + 2 = 28

Answer 66

another method for 16: \[(2+2)*(2+2)= 16\]

Answer 67

\[2^{2^{2^{2}}}=2^{2^{4}}=2^{16}=65536\]
Maybe throw in a factorial sign or two in there.
\[2^{222!}\] or \[22!^{22!}!\]
Both are incredibly larger than the first one I gave and can't be expressed by Google, apparently.

Answer 68

Fun ♦o♦