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Mathematics 13 Online
OpenStudy (lgbasallote):

LGBARIDDLE!!! 1 + 2 = 3 1 + 4 = 5 3 + 2 = 5 1 + 1 = 3 so what's 2 + 5?

OpenStudy (anonymous):

20 bucks natural log and hyperbolic secant are in the answer.... :P

hartnn (hartnn):

are u sure question is 2+5 and not 1+5 ??

OpenStudy (lgbasallote):

of course im sure...im the one who made it lol

hartnn (hartnn):

because i got 1+5 ....

OpenStudy (anonymous):

I'm getting 5, but i'm not sure xD

OpenStudy (lgbasallote):

why would you say 5??

OpenStudy (anonymous):

1+2=3, so 1=1 1+4=5, so 1=1 3+2=5, so 3=3 1+1=3, so 1=2 3+(1=2)=5, so 4=5 so 2+5 = (1+4) = 5 idk if this makes any sense, but i'm assuming 7 isn't the answer here haha xD

OpenStudy (lgbasallote):

this isn't programming.... o.O

OpenStudy (anonymous):

oh... i didn't know there was a theme to the riddle xD

OpenStudy (lgbasallote):

well...it is in the math section...

OpenStudy (anonymous):

hahaha touche sir xD

hartnn (hartnn):

if u give me 3+3 or 4+4 i can give u 2+5 all these 3 are equal.

OpenStudy (lgbasallote):

not really...

Parth (parthkohli):

I'm cool with the first three, but the fourth messes it up. :|

OpenStudy (anonymous):

6

OpenStudy (lgbasallote):

vintage lgbariddles

OpenStudy (lgbasallote):

@Omniscience no

OpenStudy (anonymous):

then its another ambiguous riddle :P

hartnn (hartnn):

any hint?

Parth (parthkohli):

2 + 5 = 7!!!

OpenStudy (lgbasallote):

@hartnn addition

hartnn (hartnn):

i did add those...1st i formed equation, a+b=c,a+d=e ....(a-1,b=2,....) but could not get b+e as single 'letter' ....am i on right track ?

OpenStudy (anonymous):

8

OpenStudy (sasogeek):

does the usual algebraic rules apply? eg. a+b=c => a=c-b ?

OpenStudy (unklerhaukus):

9

OpenStudy (lgbasallote):

still no

OpenStudy (sasogeek):

i have a feeling the answer is a decimal -_-

OpenStudy (lgbasallote):

nope. it's whole

OpenStudy (sasogeek):

well you didn't answer my earlier question lol

OpenStudy (lgbasallote):

i dont tink it's relevant

OpenStudy (sasogeek):

i think it is, i wanted to try something which might help so i need that info :) at least i have a right to know if that's possible or not :P

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