Impossible to complete ;)

Hmmmmmm I am not happy with the way Fishy is using square

I would want to use the 2 for square otherwise, what is to stop us from using any power we want

Yeah I stopped :tongue:

Reply 41

TenOfThem

18

Original post by L'Evil Fish

Yeah I stopped :tongue:

lol

did you get 19 it is annoying me

Reply 42

L'Evil Fish

18

Original post by TenOfThem

lol

did you get 19 it is annoying me

Hmmmm... He said you could used squared once :tongue:

but without that..

Reply 43

Hasufel

16

if using squares once only, cant you do for 19:

2+4^{2}+1+0

?

(or, sorry - are you trying to get it a certain way? - apologies if you are!)

Reply 44

TenOfThem

18

Original post by Hasufel

if using squares once only, cant you do for 19:

2+4^{2}+1+0

?

(or, sorry - are you trying to get it a certain way? - apologies if you are!)

You can only use square once because you only have one 2

You have used two 2s

Reply 45

Dilzo999

19

Original post by TenOfThem

lol

did you get 19 it is annoying me

I don't think it's possible :s-smilie:

. And I don't know how you'd prove it either.

Reply 46

Dilzo999

19

Original post by TenOfThem

You can only use square once because you only have one 2

You have used two 2s

In the middle of my maths lessons, I figured it out

4!-(2+1)!+\cos(0)=19

.

Reply 47

TenOfThem

18

Original post by Dilzo999

In the middle of my maths lessons, I figured it out

4!-(2+1)!+\cos(0)=19

.

Nice

I have been avoiding trig but 0! does the same :smile:

Reply 48

ExcitinglyMundane

7

Original post by Dilzo999

In the middle of my maths lessons, I figured it out

4!-(2+1)!+\cos(0)=19

.

On that note:

14 = \frac{4!}{2} +1 + cos(0)[br]15 = (4^2) -1 + 0[br]16 = 4^2 + (0\times1)[br]17 = 4^2 + 1 + 0[br]18 = 4^2 + 1 + cos(0)[br]19 = 4! - (2+1)!+\cos(0)[br]20 = 4! - (2 + 1 + cos(0))[br]21 = 4! - (2 + 1 + 0)[br]22 = 4! - (2 + (1 \times 0))[br]23 = 4! - (1 + (2 \times 0))[br]24 = 4! - ((2 + 1) \times 0)[br]25 = 4! + (1 + (2 \times 0))[br]26 = 4! + (2 + (1 \times 0))[br]27 = 4! + 2 + 1 + 0[br]

[br]28 = 4! + 2 + 1 + cos(0)[br]29 = 4! + (2+1)! - cos(0)[br]30 = 4! + (2+1)! + 0[br]31 = 4! + (2+1)! + cos(0)[br]32 = 2^{4+1} + 0[br]33 = 2^{4+1} + cos(0)[br]34 = \sum_{r=1}^{4}(r^2 + cos(0))[br]35 = \sum_{r=0}^{4}(r^2 + 1)[br]36 = (4+2)^{1 + cos(0)}[br]37 = \sum_{r=2-1}^{4}(r! + cos(0))[br]38 = \sum_{r= 2\times 0}^{4}(r! + 1))[br]39 = (\sum_{r=1}^{4}(r! + cos(0))) +2[br]40 = (\sum_{r=1}^{4}(r! + 2)) - cos(0)[br]41 = (\sum_{r=1}^{4}(r! + 2)) + 0)[br]42 = (\sum_{r=1}^{4}(r! + 2)) + cos(0))[br]43 = (\sum_{r=0}^{4}(r! + 2)) -1[br]44 = (\sum_{r=0}^{4}(r! + 2)) \times 1[br]45 = (\sum_{r=0}^{4}(r! + 2)) + 1[br]46 = (4! - 1) \times 2 + 0[br]47 = (4! - 1) \times 2 + cos(0)[br]

[br]48 = 4! \times 2 + (0 \times 1)[br]49 = 4! \times 2 + 0 + 1[br]50 = 4! \times 2 + cos(0) + 1[br]51 = 2 \times (4! +1) + cos(0)[br]52 = 2 \times (4! +1 + cos(0))[br]53 = \sum_{r=1}^{2}(4! + r + cos(0))[br]54 = \sum_{r=0}^{1}(4! + r! + 2)[br]

34 was hard!

(edited 10 years ago)

Reply 49

Dilzo999

19

Original post by ExcitinglyMundane

On that note:

14 = \frac{4!}{2} +1 + cos(0)[br]15 = (4^2) -1 + 0[br]16 = 4^2 + (0\times1)[br]17 = 4^2 + 1 + 0[br]18 = 4^2 + 1 + cos(0)[br]19 = 4! - (2+1)!+\cos(0)[br]20 = 4! - (2 + 1 + cos(0))[br]21 = 4! - (2 + 1 + 0)[br]22 = 4! - (2 + (1 \times 0))[br]23 = 4! - (1 + (2 \times 0))[br]24 = 4! - ((2 + 1) \times 0)[br]25 = 4! + (1 + (2 \times 0))[br]26 = 4! + (2 + (1 \times 0))[br]27 = 4! + 2 + 1 + 0[br]

[br]28 = 4! + 2 + 1 + cos(0)[br]29 = 4! + (2+1)! - cos(0)[br]30 = 4! + (2+1)! + 0[br]31 = 4! + (2+1)! + cos(0)[br]32 = 2^{4+1} + 0[br]33 = 2^{4+1} + cos(0)[br]34 = \sum_{r=1}^{4}(r^2 + cos(0))[br]35 = \sum_{r=0}^{4}(r^2 + 1)[br]36 = (4+2)^{1 + cos(0)}[br]37 = \sum_{r=2-1}^{4}(r! + cos(0))[br]38 = \sum_{r= 2\times 0}^{4}(r! + 1))[br]39 = (\sum_{r=1}^{4}(r! + cos(0))) +2[br]40 = (\sum_{r=1}^{4}(r! + 2)) - cos(0)[br]41 = (\sum_{r=1}^{4}(r! + 2)) + 0)[br]42 = (\sum_{r=1}^{4}(r! + 2)) + cos(0))[br]43 = (\sum_{r=0}^{4}(r! + 2)) -1[br]44 = (\sum_{r=0}^{4}(r! + 2)) \times 1[br]45 = (\sum_{r=0}^{4}(r! + 2)) + 1[br]46 = (4! - 1) \times 2 + 0[br]47 = (4! - 1) \times 2 + cos(0)[br]

[br]48 = 4! \times 2 + (0 \times 1)[br]49 = 4! \times 2 + 0 + 1[br]50 = 4! \times 2 + cos(0) + 1[br]51 = 2 \times (4! +1) + cos(0)[br]52 = 2 \times (4! +1 + cos(0))[br]53 = \sum_{r=1}^{2}(4! + r + cos(0))[br]54 = \sum_{r=0}^{1}(4! + r! + 2)[br]

34 was hard!

Wow nice use of a summation :biggrin:

.

Reply 50

ExcitinglyMundane

7

Original post by Dilzo999

Wow nice use of a summation :biggrin:

.

Thanks :-) Time for more!

[br]55 = \sum_{r=0}^{1}(4! + (r+2)! - r)[br]56 = \sum_{r=0}^{1}(4! + (r+2)!)[br]57 = \sum_{r=0}^{1}(4! + (r+2)! + r)[br]

Reply 51

benplumley

11

Original post by Dilzo999

I don't think it's possible :s-smilie:

. And I don't know how you'd prove it either.

Original post by TenOfThem

lol

did you get 19 it is annoying me

I know I'm too late now as it's already been got, but this function also evaluates to 19. I'm not happy with it because int() is only really a mathematical function in code, I've never used it in pure maths.

int(2^4+arccos(0)+arcsin(1))=19

(edited 10 years ago)

Impossible to complete ;)

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