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

• 5*5*29*1*1=725
  68-63*68-63*68-39*68-67*68-67=725
  so a=63,b=63,c=39,d=67,e=67
  
• 11 years agoHelpfull: Yes(49) No(7)
• 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+e = 299
• 11 years agoHelpfull: Yes(28) No(6)
• 232 is the answer
• 11 years agoHelpfull: Yes(11) No(17)
• On factorizing 725, we get 5,5,29,1. There are only 4 factors and we need 5 of them. The question asks for distinct numbers and does not specify whether positive or negative numbers.
  So the numbers are 39, 63,67, 69, 73.
  (68-39)(68-63)(68-67)(68-69)(68-73)= 29*5*1*-1*-5=725
  
  Hence, a=39;b=63; c=67; d=69; e= 73.
  a+b+c+d=238. That's all folks!
• 9 years agoHelpfull: Yes(10) No(1)
• here many ans. are possible. so it depends on options.
  CASE 1:
  we can write 1*1*1*25*29=725 a=b=c=67,d=43,e=39 so (a+b+c+d=244)
  case2 : 1*1*5*5*29=725 a=b=67,c=d=63,e=39 so ans will be (a+b+c+d=260)
  case3 : 1*1*1*29*25
  case4 : 1*1*29*25*1 here a=b=67, c=39,d=43, e=1 (a+b+c+d=216) many more possibilities ..
• 11 years agoHelpfull: Yes(8) No(2)
• 5*5*29*1*1=725
  68-63*68-63*68-39*68-67*68-67=725
  so a=63,b=63,c=39,d=67,e=67
  if its a+b+c+d
  ans=232
  if its a+b+c+d+e
  ans=299
  
• 11 years agoHelpfull: Yes(6) No(3)
• a+b+c+d=299
• 11 years agoHelpfull: Yes(3) No(3)
• the correct answer is 321=49+63+73+69+67 here a b c d e are different integers.
• 11 years agoHelpfull: Yes(2) No(1)
• Answer will be 39+63+63+67+67=299
• 9 years agoHelpfull: Yes(2) No(1)
• 725 factors are 5*5*29*1*1
  then putting the values in this form
  a=63 b=63 c=39 d=67 e=67
  then these value used
  a+b+c+d = 63+63+39+67 = 232
• 8 years agoHelpfull: Yes(2) No(0)
• if we factorize 725, the factors are 29,5 and 5. so a=29 b=25 c=d=e=67.
  => 29+25+67+67=214
• 11 years agoHelpfull: Yes(1) No(4)
• 63+63+39+67+67=299
• 11 years agoHelpfull: Yes(1) No(2)
• factorize 725, factors are 5,5,29 & 1.
  or in other words we can also write 5*5*29*1 or 5*5*29*1*1 therefore a+b+c+d=63+63+39+67=232
  
• 11 years agoHelpfull: Yes(1) No(1)
• (68-a)(68-b)(68-c)(68-d)(68-e) = 725
  find a+b+c+d=?
  
  how many ways is it possible?
• 8 years agoHelpfull: Yes(1) No(1)
• in the question its given that all the integers r distinct and that too only in ascending order....plzz suggest the solution
• 10 years agoHelpfull: Yes(0) No(1)
• 725 can be written as 71*5*5
  hence a should be -3, d and e can be 67 and if d and e are so then b,c must be 63,63
  by this a+b+c+d=190.
• 9 years agoHelpfull: Yes(0) No(0)
• 260
  a=67
  b=67
  c=63
  d=63
  e=39
• 9 years agoHelpfull: Yes(0) No(0)
• 725= 5*5*29*1*1
  725 = (68-63) (68-63) (68-39) (68-67) (68-67)
  A+B+C+D = 63+ 63+ 39 + 67= 232
• 9 years agoHelpfull: Yes(0) No(0)
• On factorizing 725, we get 5,5,29,1. There are only 4 factors and we need 5 of them. The question asks for distinct numbers and does not specify whether positive or negative numbers.
  So the numbers are 39, 63,67, 69, 73.
  (68-39)(68-63)(68-67)(68-69)(68-73)= 29*5*1*-1*-5=725
  
  Hence, a=39;b=63; c=67; d=69; e= 73.
  a+b+c+d=238
• 8 years agoHelpfull: Yes(0) No(0)
• Product of 5 terms equal to 1127. As all the five terms are integers, given product should be a product of 5 numbers. Now factorize 1127.
  1127 = 72 * 23 = 7 * 7 * 23
  But given that all the a, b, c, d, e are distinct. And we are getting only 3 terms with 7 repeats.
  Now the logic is, integers means positive and negative, 7 and - 7 possible and 1, - 1 also possible .
  As a,b, c, d, e are in ascending order, the factors should be in decreasing order. So (23, 7, 1, -1, -7)
  Now a = 53; b = 69; c = 75; d = 77
  
  a + b + c + d = 274.
• 6 years agoHelpfull: Yes(0) No(0)