How many no(n) r there b/w 1 & 200 such that n/2,n/3 & 2n +1/5 r all composite natural no. ?
a]0
b]1
c]2
d]3

• Ans:-2
  As we need a number which should be divisible by 2 n 3 and 2n+1 must be divisible by 5...Lcm of (2,3)=6..so reqrd no must be a multiple of 6..only 162 and 192 satisfy the condition...162/2,162/3,(2*162+1)/5=81,54,65 all of them are composite number...similarly 192 will give three composite number
• 10 years agoHelpfull: Yes(27) No(2)
• answer is 7
  12,42,72,102,132,162,192 are the seven numbers
• 10 years agoHelpfull: Yes(16) No(24)
• a composite number is a number which has a factor other than the number itself (i.e it is not a prime number) n satisfies for (6*2 = 12) , 6*7= 42, 6*12= 72, 6*17=102,6*22=132,6*27=162, 6*32=192) out of this only 162 and 192 produce numbers which are not prime.. hence ans is 2...
  
  
  ans: 2
  
• 10 years agoHelpfull: Yes(6) No(1)
• Ans: i think it should be 0
  
  no number satisfy the last condition i.e 2n+1/5
  
  
  
• 10 years agoHelpfull: Yes(3) No(12)
• answer is c)2
  162
  192
• 10 years agoHelpfull: Yes(2) No(4)
• ans should be 2 because only number 82 and 162 satisfies these expression and gives composite number
• 10 years agoHelpfull: Yes(1) No(9)
• either options or que is wrong???
  12,42,72,102,132,162,192 all 7 are possible
• 10 years agoHelpfull: Yes(0) No(6)
• friends, is dis (2n+1)/5????????
• 10 years agoHelpfull: Yes(0) No(1)
• Ans:option(d)-3, since the no has to be divisible by both 2,3& 2n+1 has to be divisible by 5 so 12,162 & 192 satisfies th cond.
• 9 years agoHelpfull: Yes(0) No(0)
• answer is 7
  12,42,72,102,132,162,192 all these number divisible by 2 and 3 and 2n+1 also divisible by 5
• 9 years agoHelpfull: Yes(0) No(0)
• If (2n+1)/5 is also an composite natural no (be the condition)
  in that case answer is '2'.
  check it for 162,192
  325/5=65, 385/5=77 . 65 & 77 are composite natural numbers
• 8 years agoHelpfull: Yes(0) No(0)
• why 2 why not 7
• 7 years agoHelpfull: Yes(0) No(0)
• Since the number should be multiple of 2 & 3, so we can say it is also a multiple of 6. Now comes (2n +1)/5 which means the last digit should be either 0 or 5. since any number multiplied by 2 gives an even number so 0 option can't be achieved at any point. Thus the 5 is the only option and to satisfy this condition the last digit from 6x which is n should have 2 in units place. As 2*..2 + 1 is a divisible option which gives 7 as an answer.
  
  6 * 2 = 12
  6 * 7 = 42
  6 * 12 = 72
  6 * 17 = 102
  6 * 22 = 132
  6 * 27 = 162
  6 * 32 = 192
• 7 years agoHelpfull: Yes(0) No(0)