4 childrens born in consective years,every year her mother calcilates the product of their ages,what is always true for that product:
a) prime no. + 1
b)square of a number -1
c) always divied by 10

• i think ans should be (b)square of a number -1
  check
  1*2*3*4=5^2=1
  2*3*4*5=11^2-1
  3*4*5*6=19^2-1
• 10 years agoHelpfull: Yes(27) No(2)
• answer should be a , prime no +1 i guess.
• 10 years agoHelpfull: Yes(7) No(11)
• ans is option b)square of a number -1 as rakesh had said earlier
  
  @vikash 119 is divisible by 7 or 17 i.e. 7*17=119
  
  & 3*4*5*6=360=361-1=19^2 -1
• 10 years agoHelpfull: Yes(5) No(0)
• Ans:B
  1*2*3*4=5^2-1
  2*3*4*5=11^2-1
  3*4*5*6=19^2-1
  4*5*6*7=29^2-1
  ....soon..
  product of the four consecutive numbers=((first*last)+1)^2-1
  ie:n*(n+1)*(n+2)*(n+3)=((n*(n+1))+1)^2-1
  
• 10 years agoHelpfull: Yes(4) No(0)
• a) prime no. +1
  case I -> 1*2*3*4=23+1=24
  case II -> 2*3*4*5=119+1=120
  case III-> 3*4*5*6=359+1=360
  case IV -> 4*5*6*7=719+1=720
  5*6*7*8=1679+1=1680
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  here 23, 119,359,719,1679 all are prime no.
  option B will be wrong because case III doesn't follow it and for option C case I doesn't follow it....
• 10 years agoHelpfull: Yes(3) No(10)
• 119 is not prime
• 10 years agoHelpfull: Yes(0) No(0)
• every prime number is 6n+-1.
  n(n+1)(n+2) is always divisible by 6.
  thus +1 would be a prime number
• 10 years agoHelpfull: Yes(0) No(0)