If (68-a)(68-b)(68-c)(68-d)(68-e)=725
then a+b+c+d=?

• ans is 232 or 236 or 260
  725=5*5*29*1*1
  725=(68-63)(68-63)(68-39)(68-67)(68-67)
  so a,b,c,d,e=63,63,39,67,67
  so A+B+C+D=232 or 236 or 260
• 10 years agoHelpfull: Yes(26) No(1)
• 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
• 10 years agoHelpfull: Yes(11) No(0)
• answer is 311
• 10 years agoHelpfull: Yes(5) No(5)
• (68-63)*(68-63)*(68-39)*(68-67)*(68-67)
  a+b+c+d=232
• 10 years agoHelpfull: Yes(2) No(0)
• Answer is 252 or 256
  725=(68-63)*(68-63)*(68-63)*(68-63)*(68-67)=5*5*5*5*1=5^4*1
  a=b=c=d=63
  a+b+c+d=63+63+63+63=252
  or
  725=(68-63)*(68-63)*(68-63)*(68-67)*(68-63)=5*5*5*1*5=5^4*1
  a=b=c=d=63
  a+b+c+d=63+63+63+67=256
• 10 years agoHelpfull: Yes(1) No(9)
• I am not understand.... plz explain clearly..
• 10 years agoHelpfull: Yes(0) No(1)
• ans:252
  here a=63,b=63,c=63,d=63 and e=68
• 10 years agoHelpfull: Yes(0) No(1)
• 725/5=145 A=5 (5*5*29*1=725) A=68-5=63
  145/5=29 B=5 B=68-5=63
  29/29=1 C=29 C=68-29=39 63+63+39+67=232
  1/1=1 D=1 D=68-1=67 ANSWER IS 232
  (68-63)*(68-63)*(68-39)*(68-67)=725
  
• 10 years agoHelpfull: Yes(0) No(0)
• 1+1+1+43=46
• 10 years agoHelpfull: Yes(0) No(1)
• 67+67+67+39=240
• 10 years agoHelpfull: Yes(0) No(0)