a,b,c,d,e are distinct integer values in ascending order and (89-a)*(89-b)*(89-c)*(89-d)*(89-e)=4961. find a+b+c+d=?

• ans = 100 +90 +88 +78 +48 =404
  soln : take factors of 4961
  we get 11 *11 *41
  therefore eqn will become = (89-100)*(89-90)*(89-88)*(89-78)*(89-48)= (-11)*11*41*1*(-1)=4691
• 11 years agoHelpfull: Yes(46) No(9)
• a,b,c,d,e can get any of these values 100,78,48,88,90
  then 11*(-11)*41*1*(-1)=4961
  depending upon the options we can get the value of a+b+c+d...
• 11 years agoHelpfull: Yes(10) No(8)
• factors of 4961 are -1*-1*11*11*41...to get a,b,c,d,e in increasing order we want 89-a,89-b,89-c,89-d,89-e in decreasing order,just equate them with 41,11,11,-1,-1 respectively and the value will be a=48'b=78,c=78,d=90,e=90, a+b+c+d=294.
• 11 years agoHelpfull: Yes(10) No(7)
• 48+78+88+90+100=404
  4961=11*-11*41*-1*1
• 11 years agoHelpfull: Yes(5) No(2)
• if you follow the 1st fw terms ypu will understand its sum(n.n!)= (n+1)!-1
  so its 2014!-1
• 11 years agoHelpfull: Yes(3) No(8)
• (89-a)*(89-b)*(89-c)*(89-d)*(89-e)= 41*11*1*-1*-11
  a= 48,b=78,c=88,d= 90,e=100
  a+b+c+d+e= (48+78+88+90+100)=404
• 11 years agoHelpfull: Yes(3) No(2)
• ans=90+100+88+78=356
• 9 years agoHelpfull: Yes(0) No(1)
• here a
• 8 years agoHelpfull: Yes(0) No(0)