If 1535/1038 = a+1/[b+1/{c+(d+1/e)}]. Find a*b*c*d*e.

• by expanding 1535/1038 make form like a+1/[b+1/{c+(d+1/e)}
  we get
  a=1, b=2,c=11,d=3,e=13/5
• 8 years agoHelpfull: Yes(16) No(18)
• by expanding 1535/1038 make form like a+1/[b+1/{c+(d+1/e)}
  =1+(497/1038) now a=1
  =1+{1/(1038/497)}
  =1+{1/(2+42/497)}
  =1+{1/(2+1/(497/42)} now b= 2 similarly expand
  c=11 d= 1 e=34/8
  finally a*b*b*c*d*e
  1*2*11*1*34/8
  187/2
  
• 8 years agoHelpfull: Yes(8) No(5)
• if the above eqn is like 1535/1038 = a+1/[b+1/{c+1/(d+1/e)}]. Find a*b*c*d*e then it can be solved. I think given form is not correct
• 8 years agoHelpfull: Yes(6) No(5)
• @Rahul Rofl... this was the worst comment I had read on this site...
• 8 years agoHelpfull: Yes(6) No(6)
• a=1,b=2,c=11,d=3,e=2.Ans :132
  
  1st 1535/1038=1+497/1038 i.e a=1
  2nd 1038/497=2+44/497 i.e b=2
  3rd 497/44=11+13/44 i.e c=3
  4th 44/13=3+5/13 i.e d=3
  5th 13/5=2+3/5 i.e e=2
• 8 years agoHelpfull: Yes(5) No(6)
• Vijay if u do 1038/497 then the ans is 2+44/497 not 2+42/497. And from where u get the value of d & e can u plz explain.
• 8 years agoHelpfull: Yes(3) No(2)
• hi pooja u r looking brilliant , will you please help me in cracking my tcs written test
• 8 years agoHelpfull: Yes(2) No(50)
• yes i m agree with you pooja
• 8 years agoHelpfull: Yes(0) No(2)
• I also in favor of Pooja sow regarding this question.
• 8 years agoHelpfull: Yes(0) No(1)
• Rahul btw its a complement or comment? :p
  Pooja u cleared tcs?
• 8 years agoHelpfull: Yes(0) No(1)
• a=1, b=2,c=11,d=3,e=13/5
  but how we get? explain this.
• 8 years agoHelpfull: Yes(0) No(1)
• 132 is the answer
• 8 years agoHelpfull: Yes(0) No(1)
• so what is the exact solution
• 8 years agoHelpfull: Yes(0) No(1)
• Right question is 3^48+3^1996+3^3943+3^3n
  so check the difference b/w to power 1996-48=1948.
  again check then we will get 3943-1996=1947.
  so here we can see that the difference b/w 1st two power and 2nd two power decreased by 1.
  so we can write for 3rd power i.e.
  3n-3943=1946
  n=1963 answer
• 4 years agoHelpfull: Yes(0) No(0)