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The number of positive number which divides either 2700 or 9000 are _________?

 i am getting 20
in Combinatory
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72 ???
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i am getting 57. whats the answer given.
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I am getting 30 :P
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Done by mutual inclusion exclusion principle :
number dividing X==> n(x)
n(2700 or 9000)=n(2700)+n(9000)-n(2700 and 9000)

n(2700)=(3*3*3) X (2*2) x (5*5)
factors of 2700 =4*3*3 =36

 9000 = (3*3)x(2*2*2)x(5*5*5)

factors of 9000= 3*4*4=48
Now we find numbers which divide both 2700 and 9000
which is the product of common factors which are (3*3)x(2*2)x(5*5)
so total factors= 3*3*3=27
so total answer=36+48-27 =57
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If it's either 2700 or 9000 (but not both) $\rightarrow 36+48- (2*27)=30$
If it's  2700 or 9000 or both  $\rightarrow 36+48- 27=57$
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I did mistake in calculation.
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yes , answer is 57 only;

okay ! i got it;

my mistake was while finding numbers which divide both 2700 & 9000 , i was finding LCM;

okay so the common factors are to be calculated!!
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@Gate Fever, but according to wording "either or" correct answer will be 30 as @Soumya29 did.

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@

If it's either $2700$ $(or)$ $9000$ (but not both)$ →36+48−$(2∗27) $=30??$

How did you subtract $2\times27?$

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IN english language where

A OR B MEANS EITHER A OR B  BUT NOT BOTH

BUT IN COMPUTER SCIENCE

A OR B MEANS EITHER A OR B THAT IS if one of the two or both of them are true,then statement is true;

although am not also sure;

moreover solution is of madeeasy , we cant rely on that too blindly.

 lets see what others say!! till then wait and consider 57 as correct answer!!
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@Lakshman Patel RJIT

it is symmetric diiff formula

$n(A\Delta B)=n(A)+n(B)-2n(A\cap B)$

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@Lakshman Patel RJIT, Draw the Venn diagram you will get it :)
@Shubhgupta 
@Gate Fever I am also confused in this "either or" thing.
@Mk Utkarsh please check this once. 

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Case $(1):$Either $A$ $(or)$ $B$  but not both

 Case $(2):$Either $A$ $(or)$ $B$  $(or)$ both

                                  $(OR)$

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