A natural number has exactly 10 divisors including 1 and itself . how many distinct prime factors can this natural number have

(a) either 1 or 2 (b)1 or 3 (c)either 2 or 3 (d) either 1, 2 or3

• (a) either 1 or 2
  
  explanation- Any natural number is represented by its prime factors (eg. 2^a * 3^b * 5^c * ..... and so on). The number of factors or divisors will be the product of all the powers after incremented by 1 like (a+1)*(b+1)*(c+1)*..and so on.
  So, in the question we got 10, now 10 can be a result if either a single power or product of 2 powers only and not more than that.
  So, here
  Case 1: (a+1)=10
  Case 2: (a+1)(b+1) = 10.
  so the answer is either 1 or 2.
• 8 years agoHelpfull: Yes(25) No(0)
• either 1 or 2
  because 10 can be breaked as either 1*10 or 2*5
• 8 years agoHelpfull: Yes(2) No(4)
• a) either 1 or 2
  As it is given the number has exactly 10 divisors including 1 and itself. so the number may be 2^9,3^9,5^9...any prime number to the power 9 will give 10 divisor.so only 1 prime factor.
  Also, the number 2^4*5^1, 3^4*5^1,....will give exactly 10 divisor which includes only 2 prime factors for any case i.e (2,3)(2,5)(3,5)....so answer will be either 1 or 2
• 8 years agoHelpfull: Yes(2) No(1)
• (c) either 2 or 3
  
  Exp- let take a example of that Natural no. for this Take the LCM of no:- 1,2,3,4,5,6,7,8,9
  = 2520 = (2^3 * 3^2 * 5 * 7)
  Now prime factor :- 5, 7 or 5,7 or any one we can take
• 8 years agoHelpfull: Yes(0) No(4)
• (C) either 2 or 3
• 8 years agoHelpfull: Yes(0) No(3)