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Maths problem

loll1es Avatar
6y, 8m agoPosted 6 years, 8 months ago
Whats is the greatest number that can be achieved using these three numbers

1, 2, 3

I am almost certain a number greater than 6 can be achieved, I'm thinking the square root divided by one of the numbers, but cant work it out
loll1es Avatar
6y, 8m agoPosted 6 years, 8 months ago
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#1
3+1 x 2
#2
1+3x2=8
#3
3/(1/(2^2)) = 12

2/(1/(3^2)) = 18
#4
321 Lol!
#5
kungfu
3+1 x 2


= 8

(2+1) x 3 = 9.

No idea if anything over that though.
#6
kungfu
3+1 x 2


+1 :thumbsup:
#7
2+1x3=9
#8
321 add up:-D
#9
r3tract
3/(sqrt(1/2)) = 12


Is that 3 divided by (1 divided by 2 sqrt)
#10
2 divided by the square root of 1/3 =18?
#11
3 to the power of 2 +1?
#12
choc1969
2+1x3=9


Thanks babes x
#13
chesso
2 divided by the square root of 1/3 =18?


Thats 2 sums on one number, if you get what I mean.

You can have 1 divided by 3 to get 1/3 and then squareroot the result, you are only allowed one sum per number, else I could just keep adding 3 to 3 until I feel like I want to stop
#14
awoodhall2003
3 to the power of 2 +1?


thats 10
#15
loll1es
thats 10


thus a greater number then 6.
#16
3 ^ (2+1) = 27 :)
#17
awoodhall2003
thus a greater number then 6.


yeah but there are already results up to 18 (might not be correct though)
#18
linuxlinks
3 ^ (2+1) = 27 :)

Yep, that's the best you can get I think
#19
2+1 = 3
3^3
= 27
#20
linuxlinks
3 ^ (2+1) = 27 :)


+1

not 27+1 lol but 27 is the highest I believe.
#21
loll1es
yeah but there are already results up to 18 (might not be correct though)

Sorry couldnt see the others when i typed it.

linuxlinks
3 ^ (2+1) = 27 :)


baffledsalmon
2+1 = 3
3^3
= 27

as above...

But...2 to the power of (1+3)
#22
awoodhall2003


But...2 to the power of (1+3)


That's a mere 16 though :)
#23
baffledsalmon
2+1 = 3
3^3
= 27


what is ^ mean? :?
#24
loll1es
what is ^ mean? :?

to the power of
#25
linuxlinks
That's a mere 16 though :)

throwing alternatives, the highest I agree with, i cant think of anything better.
loll1es
what is ^ mean? :?

to the power of.
#26
linuxlinks
That's a mere 16 though :)


yep.. and 27 is greater than 16 last time I checked too. Unless its a trick question... 27 is correct.
#27
I think 27 is the highest
#28
5nowman
yep.. and 27 is greater than 16 last time I checked too. Unless its a trick question... 27 is correct.


Yeah, such as if we were allowed to use ! (factorial :)
#29
31 to the power of 2 = 961

Does this count? :thinking:
#30
ScreechMF
31 to the power of 2 = 961

Does this count? :thinking:


No
#31
If that's allowed then 21 to the power of 3=9261
#32
ScreechMF
31 to the power of 2 = 961

Does this count? :thinking:


Nope
#33
(3^(2+1))! = 10888869450418352160768000000

Quite a resounding victory I think! :p
#34
InfernoZeus
(3^(2+1))! = 10888869450418352160768000000

Quite a resounding victory I think! :p


Think your calculator is broke, cos mine makes that 27 :lol:
#35
loll1es
Think your calculator is broke, cos mine makes that 27 :lol:

Then you forgot to use the factorial button :w00t:

(3^(2+1))! = 27! = 10888869450418352160768000000
#36
! isn't permitted
#37
linuxlinks
! isn't permitted


Why not? :?
#38
InfernoZeus
Then you forgot to use the factorial button :w00t:

(3^(2+1))! = 27! = 10888869450418352160768000000


thats using 2 sums on one number, not permitted, sorry
#39
InfernoZeus
Why not? :?


cos I said so, and I make the rules.....ha ha!!
#40
loll1es
thats using 2 sums on one number, not permitted, sorry

Not really, I make one sumation of 1 and 2, then the product of the result with 3, and then the factorial of the product.

No different to summing 1 and 2, and then raising the result to the power 3.

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