Question

Can you answer this....

Answer 1

Using only two 2's and any combination of mathematical signs, symbols, and functions and make 5.

Answer 2

2 + 2 + 1?

Answer 3

only two 2's

Answer 4

I only have two 2's

Answer 5

no other numbers

Answer 6

hint : there is a square root involved

Answer 7

well to answer that we need to know all the mathematical signs, symbols, and whatever. There is hundreds however and I or anyone else can't simply know them all. But if call the number of distinct mathematical signs, symbols and functions 'n', then the number of combinations of two 2s and 3 signs/symbols/functions would be given by:\[\bf \left(\begin{matrix}n \\ 3\end{matrix}\right)=K\], and the answer would be some integer K. Now here I'm assuming that by 'combination' you do in fact mean 'combination' which means that order doesn't matter. @kelliegirl33

Answer 8

Are mathematical constants allowed?

Answer 9

@kelliegirl33 And where are you getting the square root from lol

Answer 10

The answer has only two two's....no letters are involved

Answer 11

No letters are involved? Doesn't that eliminate almost all mathematical functions?

Answer 12

it is the square root of something.....using only two 2's

Answer 13

Well, you established that the only numbers allowed are two 2s, so by logic it must be square root 2.

Answer 14

is it...\(\large (\sqrt{5})^2\)

Answer 15

two 2's...no other numbers.....negatives and positives are involved

Answer 16

omg lol.

Answer 17

its tricky....until you see it

Answer 18

there is a decimal involved

Answer 19

trig functions work?

Answer 20

don't need trig functions

Answer 21

i was thinking of the unit circle and square roots haha...

Answer 22

there is " to the power of " sign ^

Answer 23

\[\csc^2(arccotangent 2)\]

Answer 24

not the answer given

Answer 25

ahhhh he picked up on my method!

Answer 26

lol...wait till you see it......not that hard

Answer 27

I actually didn't come up with that solution and can't take credit.

Answer 28

@kelliegirl33 All this time I was thinking this was a combinatorics problem but seems like its not? I feel as if you're not asking the question correctly which is why I'm misinterpreting it.

Answer 29

I am asking it like it is printed on this page....

Answer 30

Can you type it in verbatim and fully one more time?

Answer 31

I agree @genius12 the presentation of this problem was very below par.

Answer 32

Seems like theres a new rule every 2 minutes

Answer 33

wait a minute,this question was asked a year ago on OS lol.

Answer 34

http://openstudy.com/study#/updates/4f8109d1e4b0505bf0831bc0

Answer 35

wat lolol

Answer 36

of course.....Using only two 2's and any combination of mathematical signs, symbols and functions , can you make 5. That is exactly like it is printed

Answer 37

Tell the person who printed it to learn how to be less ambiguous.

Answer 38

lol....

Answer 39

OH so it's adding/subtracting/dividing/multiplying and other operations sort of thing. You are to use two 2s and a combination of operators to get to 5. Now I get rofl. I though we were just making combinations of two 2s and any 3 mathematical symbols such that there is 5 total characters.

Answer 40

The correct answer is on that previous copy of this question if you want it.

Answer 41

square root of something....there is a decimal and a power up sign ^

Answer 42

You could do the following:\[\bf 2^2 + 2^0 = 5\]1 symbol and two 2s. Done. @kelliegirl33 @vinnv226

Answer 43

Unless the 2 in the exponent counts as a 2 too then that would be dumb lol...

Answer 44

it counts

Answer 45

I still like this solution better:
\[\csc^{2} (arccot 2)=1+\cot^{2}(arccot 2)=1+2^{2}=5\]

Answer 46

there is also a negative and a decimal involved

Answer 47

lol and someone else posted this a year before the year posted.... hahaha

Answer 48

what was the answer on that one ?

Answer 49

Someone on that page came up with the solution \[\bf \lceil \sqrt{22} \rceil\]and imo it works fine; two 2s and mathematical functions

Answer 50

one problem..it does not equal 5

Answer 51

yeah...the "ceiling" would be the max number it could reach..which in this case sqrt(22) approx = 4.69 rounded up to 5...

Answer 52

hence "ceiling"

Answer 53

do you want the answer ?

Answer 54

Yep.

Answer 55

That answer works too. So far we have 2 solutions.

Answer 56

I imagine there are more though.

Answer 57

here is the answer given.....
sq rt of .2^-2

Answer 58

There are an infinite number of solutions....just depends on what integers you use.

Answer 59

Eh. Thats a third solution but I think the first two are better.

Answer 60

lol.....I didn't get it the first time I looked at it..

Answer 61

0.2 = 1/5 1^(-2) / 5^(-2) = 1/(1/5^2) = 25. That does not work lol. @kelliegirl33

Answer 62

No genius......

Answer 63

sq rt of (.2 ^ -2)

Answer 64

oh sqrt lol

Answer 65

Ya but the solutions mentioned before are better

Answer 66

I agree

Answer 67

\[\large (\sqrt{0.2})^{-2} = \cfrac{1}{(\sqrt{0.2})^2}=5\]

Answer 68

If you agree why were you disagreeing before?.... lmao.

Answer 69

Doesn't the square root technically use a two since it's the 2nd root? If you just put the square root sign, it's generally accepted as square root but still, to be clear you'd put a 2.

Answer 70

I wasn't disagreeing...I was just saying that it wasn't the answer given

Answer 71

Ah... ok.

Answer 72

good job peeps....you are all very smart

Answer 73

I just figured another one:\[\bf \lceil \sqrt{(2*2)!} \rceil\]Also works. @kelliegirl33 @vinnv226 @Jhannybean

Answer 74

yeah, another ceiling function.

Answer 75

Yep, that works too.

Answer 76

Same thing you've written but (2^2) also works.

Answer 77

You could also make it 2 + 2 instead of 2 x 2 or you could make it 2^2

Answer 78

:P

Answer 79

oh yeah...your right

Answer 80

We're up to 7 unique solutions then

Answer 81

this is one easy riddle. throw something harder at us @kelliegirl33 lol

Answer 82

I will have to.....you people are too smart

Answer 83

\[\bf \lceil \sqrt{(2*2)!} \rceil \approx 4.89\] round up.

Answer 84

Plus you could put:
\[ceiling \sqrt{(2! + 2!)!}\] and do above, which would put us at 11 solutions.

Answer 85

lol @vinnv226 you're now just modifying what I did but that still works =P

Answer 86

Thats collaboration!

Answer 87

hence my post in the beginnning,there are infinite solutions, depending on the integers used.

Answer 88

thats ok...I WILL find one you can't do......lol

Answer 89

You could also do:\[\bf \lceil \sqrt{(2!^{2!})!} \rceil\]

Answer 90

<_< thats too complicated for me lol.

Answer 91

The ceiling function is beautiful isn't it?

Answer 92

lol...

Answer 93

thanks peeps.......it was fun....ceiling function is good

Answer 94

it's actually ugly because its a bunch of straight lines but its definitely helpful

Answer 95

Wait I got another one let me type it up

Answer 96

where erf(x) is the error function