Q. The multiplication of three consecutive number is 2730 than the sum of the three no. is

Option
a) 39
b) 42
c) 30
d) None of these

• factors of 2730 = 2 x 5 x 3 x 13 x 7
  = 13 x 7 x 2 x 5 x 3
  = 13 x 14 x 15
  so numbers are = 13, 14 ,15
  addition = 13 + 14 + 15
  =42 ans
• 11 years agoHelpfull: Yes(35) No(0)
• b)42,
  (x-1)*x*(x+1)=2730
  =>x=14
  =>numbers are 13,14,15
  their sum is 42
• 11 years agoHelpfull: Yes(12) No(7)
• if 3 numbers are consequtive then the multiplication will be nearest to their average cube ie(14*14*14=2744), nearest cube.
  hence numbers are 13,14,15..
• 11 years agoHelpfull: Yes(9) No(1)
• lets say the numbers are x,(x+1),(x+2).then x*(x+1*(x+2)=2730.
  sum of these 3 numbers => 3x+3.
  by trail & error method option 1)3x+3=39 =>x=12 but 12*13*14 is not equal to 2730
  if we take 3x+3= 42 =>x=13 then 13*14*15=2730.
  So Ans is b)42
• 10 years agoHelpfull: Yes(3) No(3)
• LCM of 2730
  You will find the factors.
  Now Take the highest number make it into Sequence Like as per consecutive numbers
  In this case Factors are 2 3 5 7 13
  Now if we take
  7 * 2 = 14 (i.e. 13+1)
  Again
  5 * 3 = 15 (i.e. 13+2)
  
  therefore, 13 14 15 are the numbers
  For cross verification you can multiply them.
  
  Therefore 13+14+15 = 42. Is the Answer
• 7 years agoHelpfull: Yes(1) No(0)
• answer is 42
• 9 years agoHelpfull: Yes(0) No(1)
• take no. as (n-1), n, (n+1)
  product =n(n-1)(n+1)= n^3 -1
  n^3 -1= 2730
  solving we get n as 14, -7+12i, -7-12i
  sum = 0
• 8 years agoHelpfull: Yes(0) No(0)
• factors of 2730 are - 2*5*3*7*13 = 2730 now they are asking consecutive no. you can see the pattern like 13,7*2=14 and 5*3 = 15 so when we add we get 13+14+15 = 42
• 8 years agoHelpfull: Yes(0) No(0)
• consecutive no's are 13,14,15
  13*14*15=2730
  sum=13+14+15=42
• 7 years agoHelpfull: Yes(0) No(0)
• factor of 2730=2*5*3*13*7
  13*14*15
  sum of these no =42
• 7 years agoHelpfull: Yes(0) No(0)
• we need to check from 10x11x12 ....... so we arrive at 13x14x15=2730
  so 13+14+15=42
  so ANS: b) 42
• 5 years agoHelpfull: Yes(0) No(0)