(68­a)(68­b)(68­c)(68­d)(68­e) = 725 ? Find a+b+c+d= ?

• solution is posible in this case also take a=b=5/68 c=29/68 d=1/68 e=1/68 then a+b+c+d=40/68
• 9 years agoHelpfull: Yes(8) No(0)
• question is wrong
  it should be
  (68-a)(68-b)(68-c)(68-d)(68-e)
  and the ans is 29*5*5*1*1
  68-a=29 => a=39
  68-b=5 => b=63
  68-c=5 => c=63
  68-d=1 => d=67
  68-e=1 => e=67
  a+b+c+d=232
• 9 years agoHelpfull: Yes(7) No(0)
• how the ans is 232
• 9 years agoHelpfull: Yes(0) No(0)
• plzzz
  explain this question
• 9 years agoHelpfull: Yes(0) No(0)
• fisrt take out the factor of 725 by using factorization. the factor is 5,5,29. but we five variables a,b,c,d,e, so take 5 as -5 becoz every no exist in +- form on number line. also arrange in ascending order, then (29,5,-5,1,-1). now compare and find abcde ie a=39,b=63,c=73,d=67,e=69. then find a+b+c+d=39+63+73+67=242. hence 232 is wrong.
  
• 9 years agoHelpfull: Yes(0) No(0)