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Q.)What is the sum of all the even factors of 2160?

(i)10800

(ii)3600

(iii)7200

(iv)14400

=>

2160 = $5*{ 2 }^{ 4 }*{ 3 }^{ 3 } $

Sum of all the even factors =${ (5 }^{ 0 }+{ 5 }^{ 1 })({ 2 }^{ 1 }+{ 2 }^{ 1 }+{ 2 }^{ 1 }+{ 2 }^{ 1 })({ 3 }^{ 0 }+{ 3 }^{ 1 }+{ 3 }^{ 2 }+{ 3 }^{ 3 })$

=$(\frac { { 5 }^{ 2 }-1 }{ 5-1 } )(\frac { { 2(2 }^{ 4 }-1) }{ 2-1 } )(\frac { { 3 }^{ 4 }-1 }{ 3-1 } )$

=6*30*40 =7200.

As the sum is to be found out only for even factors, so 2 always has to be included. So the sum started with ${ 2 }^{1 }$.

Option(iii)