(79-a)(79-b)(79-c)(79-d)(79-e)=2299, where a,b,c,d,e are distinct and in ascending order. So find a+b+c+d

• ans:286
  since 2299 is divisible by 11
  it can be written as : 19*11*1*-1*-11
  thus,put a=60,b=68,c=78,d=80,e=90 which satisfies above condition (79-a)(79-b)(79-c)(79-d)(79-e)=2299
  =>a+b+c+d=286
• 11 years agoHelpfull: Yes(54) No(8)
• Since we can write 2299 = 19*11*11
  and a,b,c,d are in ascending order so it may be as 19*11*1*(-1)*(-11) = 2299
  So the values satisfying equation is (79-60)(79-68)(79-78)(79-80)(79-90)=2299
  so values are a=60,b=68,c=78,d=80,e=90
  Thus a+b+c+d+e = 376
  
• 11 years agoHelpfull: Yes(11) No(5)
• madhu pericherla..can u give the option of the above question?
  
• 11 years agoHelpfull: Yes(4) No(0)
• The no. of ways in which a selection of 4 letters can be made from the letters of the word INFINITE. a) 32 b) 15 c) 37 d) 22
• 11 years agoHelpfull: Yes(2) No(7)