a natural no has exactly 10 divisors including 1 and itself.how many distinct prime factors can this no have?

• The number can have two distinct prime numbers.
  10 can be written as 1*10 and 5*2
  By rule (a+1)(b+1) should give 10
  So (4+1)(1+1)=10.
  thus two distinct prime numbers.
• 11 years agoHelpfull: Yes(10) No(12)
• actually options for these question are:
  a)only 1 b)either 1 or 2 c)2 prime factors d) either 1 or 2 or 3 prime factors
  write sequence such that you invove minimum number of prime numbers in the sequence:
  involve only 1 prime number i.e., 2 and write sequence upto 10 number in the following procedure:
  ->1 2 4 8(2*4) 16(8*2) 32(16*2) 64(16*4) 128 256 512 1024:
  number is 1024 having 10factors including 1 and itself including 1 prime factor i.e., 2
  now involve 2 prime numbers 2 and 3 by u r self in the sequence
  ->1 2 3 6 12 18 24 36 54 72:number is 72 and no of primes is 2(i.e., 2 and 3)
  similarly another sequence in the same way
  -> 1 2 3 5 10 15 20 30 40 50 :number is 50 and no of prime factors is 3 i.e., 1,2,3
  so from all the above conditions the answer is d) either 1 or 2 or 3
• 11 years agoHelpfull: Yes(5) No(11)
• 10=10*1
  in this case a no can have one single prime no as a divisor.
  eg 512=2^9
  this will have exactly(9+1)=10 divisors and only 2 as a prime no divisor.
  10=5*2
  in this case the number may have 2 prime no as divisors.
  so the answer is- 1 or 2
• 11 years agoHelpfull: Yes(5) No(1)
• what is the xaact soln......@ravi,karthiga???
• 11 years agoHelpfull: Yes(0) No(0)