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

• by factorizing 725, we get prime factors 5,5 and 29 i.e.,
  725=5*5*29
  therefore we can assume that the expressions enclosed within each pair of
  parentheses results into two 5's and a single 29. The remaining to expressions are bound to result into two 1's.
  like let's say,
  (68-a)=5
  (68-b)=5
  (68-c)=29
  (68-d)=1 and
  (68-e)=1
  for the above to be true, value of
  a=63
  b=63
  c=38
  d=67
  e=67
  therefore
  a+b+c+d=38+63+63+67
  =231
• 9 years agoHelpfull: Yes(19) No(11)
• The L.C.M of 725 = 5 * 5 * 29 * 1 * 1
  we can form above things like this b/c :-
  (68-a)(68-b)(68-c)(68-d)(68-e) => 5 * 5 * 29 * 1 * 1
  As,
  (68-a)=5 ; so, a=63;
  (68-b)=5 ; so, b=63;
  (68-c)=29; so, c=39;
  (68-d)=1 ; so, d=67;
  (68-e)=1 ; so, e=67;
  So we can write ,
  (a+b+c+d)=(63+63+39+67)= 232 ;
  
  N.B : This Math solution can be different if anyone consider
  (68-a) = 1 or 5 or 29
  (68-b) = 1 or 5 or 29
  (68-c) = 1 or 5 or 29
  (68-d) = 1 or 5 or 29
  (68-e) = 1 or 5 or 29
  That why it will confuse every one .
  As it is not given no relation between a,b,c,d,e ( I mean a>b>c>d>e like
  this ).
• 9 years agoHelpfull: Yes(11) No(2)
• factor or 725=5*5*29
  68-a=5
  68-b=5
  68-c=29
  68-d=1
  68-e=1
  but sequence can be change so ans will be,,63+63+39+67=232 and 63+39+67+67=236,,,and 63+63+67+67=260,,, go to option
• 9 years agoHelpfull: Yes(2) No(0)
• this question is not complete. According to given informations it can have multiple answers.
• 9 years agoHelpfull: Yes(1) No(3)
• 725 = 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+e = 299
• 9 years agoHelpfull: Yes(1) No(0)
• 725 factors are 29x5x5x1x1....
  so as we equal the 68-a=29 then a=39
  consequently b=63,c=63,d=67,e=67
  so a+b+c+d=232
• 9 years agoHelpfull: Yes(0) No(0)
• we have 725= 5x5x29
  hence we can have (68-67)(68-67)(68-63)(68-63)(68-39)=725
  hence a+b+c+d+e= 67+67+5+5+29=173
• 9 years agoHelpfull: Yes(0) No(1)
• we have 725= 5x5x29
  hence we can have (68-67)(68-67)(68-63)(68-63)(68-39)=725
  hence a+b+c+d+e= 67+67+63+63+39=299
• 9 years agoHelpfull: Yes(0) No(0)
• first of all factorize 725- >> 29*5*5*1*1
  now try to manipulate all the factors in the form of the equation
  (68-63)*(68-63)*(68-39)*(68-67)*(68-67)
  a =63
  b=63
  c=39
  d=67
  e=67
  now a+b+c+d = 232
• 9 years agoHelpfull: Yes(0) No(0)
• In actual question, there are 5 elements and they are distinct, i.e. a,b,c,d,e & a!=b!=c!=d!=e, it can be solved like, 725=29*5*(-5)*1*(-1)
  which makes the values 63+73+69+67+39 = 311
• 9 years agoHelpfull: Yes(0) No(0)
• 725=5*5*29
  68-5=67
  so a=b=67
  68-29=39 so c=39
  d=67,e=67 by using this a+b+c+d=232
• 9 years agoHelpfull: Yes(0) No(0)
• factorizing of 725,we get prime factors 5,5 and 29 i.e.,
  725=5*5*29
  (68-a)=5
  (68-b)=5
  (68-c)=29
  (68-d)=1
  (68-e)=1
  a=63
  b=63
  c=39
  d=67
  e=67
  a+b+c+d
  63+63+39+67=232
• 9 years agoHelpfull: Yes(0) No(0)
• 
  Ans: 311
  
  (68-a)(68-b)(68-c)(68-d)(68-e) = 725
  factors of 725=5*5*29
  68-a=5 ==>a=63
  As a, b, c, d can not have similar value,
  so, 68-b=-5 b=73
  68-c=29 ==> c=39
  68-d=1 ==> d=67
  68-e=-1 ==> e=69
  Hence sum
  a+b+c+d= 63+73+39+67+69=311
• 9 years agoHelpfull: Yes(0) No(0)