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by anshumishra » Wed Jan 05, 2011 3:58 pm
Night reader wrote:How many even, positive divisors does 540 have?

A. 6
B. 8
C. 12
D. 15
E. 16
540 = 2^2*3^3*5
No. of even divisors= Total divisors- odd divisors= (2+1)*(3+1)*(1+1) - (3+1)*(1+1) = 24 - 8 = 16.

E
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by gmat7202011 » Wed Jan 05, 2011 5:38 pm
540 = 2^2*3^3*5

Or directly total number of even divisors = (2) * (3+1) * (1+1) = 2 * 4 * 2 = 16

In order to get to the even divisors, we eliminate the 2 raised to 0 factor, hence 1 less.
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by anshumishra » Wed Jan 05, 2011 7:23 pm
gmat7202011 wrote:540 = 2^2*3^3*5

Or directly total number of even divisors = (2) * (3+1) * (1+1) = 2 * 4 * 2 = 16

In order to get to the even divisors, we eliminate the 2 raised to 0 factor, hence 1 less.
Right. Nice !
Multiplication by any factor of 2^n (except n=0) makes sure that any thing else multiplied becomes even.
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by Anurag@Gurome » Wed Jan 05, 2011 10:53 pm
Night reader wrote:How many even, positive divisors does 540 have?

A. 6
B. 8
C. 12
D. 15
E. 16
540 = (2²)*(3³)*(5)

In any even divisor of 540, there must be at least one 2 but there may or may not be any 3 or 5.

Number of such divisor = (Number of way to choose at least one 2 out of two 2)*(Number of ways to select any number of 3's out of three 3)*(Number of ways to select any number of 5 out of one 5) = 2*(3 + 1)*(1 + 1) = 16

The correct answer is E.
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