If (68-a)(68-b)(68-c)(68-d)(68-e)=725
then a+b+c+d=?	Choose the correct option:
A) 231  B) 231  C) 233  D) 234  E) 235
we will first break 725 in prime factor=(5*5*29*1*1) a/c to whatever factors is here   then we we will compare like   (68-63)(68-63)(68-39)(68-67)(68-67)i.e.5*5*29*1*1   a=63   b=63   c=39   d=67   and now sum is =a+ b+ c+ d=232   ANSWER:231
<result>B</result>
If (68-a)(68-b)(68-c)(68-d)(68-e)=725
then a+b+c+d=?