A natural number has 43 divisors (excluding 1 and itself) . What is the maximum number of distinct prime factors it can have ?

a) 2
b) 3
c) 5
d) None of these

• no. of total divisors=43+2=45
  If N=a^p*b^q*c^r.... then the number of factors of N = (p+1)(q+1)(r+1) ....
  45=3*3*5
  distinct prime factors are 3 and 5 therefore answer is A)2
• 10 years agoHelpfull: Yes(36) No(13)
• no. of total divisors=43+2=45
  If N=a^p*b^q*c^r.... then the number of factors of N = (p+1)(q+1)(r+1) ....
  45=3*3*5
  so there can be maxm 3 distinct prime factor
• 10 years agoHelpfull: Yes(24) No(15)
• plz explain why 43+2??
  
• 10 years agoHelpfull: Yes(5) No(1)
• once observe the question he has given excluding 1 and num.so,the no. of divisors are 43+2@adityesh mohanty
• 10 years agoHelpfull: Yes(2) No(0)
• its asking distinct prime factors so answers is 2
• 10 years agoHelpfull: Yes(1) No(0)
• correct answer is b i.e. 3
• 10 years agoHelpfull: Yes(0) No(1)
• how to get 3*3*5 explane
  
• 10 years agoHelpfull: Yes(0) No(0)
• number 43 has 3 factors but 3 is repetead twice. so there is no use.therefore 43 has 2 factors
  
• 7 years agoHelpfull: Yes(0) No(1)