Elitmus
Exam
Numerical Ability
Number System
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. Either 1 or 3
C. Either 2 or 3
D. Either 1,2 or 3
Read Solution (Total 5)
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- Either 1 or 2
Reason : since given maximum factors is 10...this is obtained only by (9+1) or (4+1)*(1+1)
Since we know that for p1^a,p2^b.... are factors ,p1 and p2 are prime...then total factors are (a+1)*(a+2)....
by this way we can't split 10 not more than 2 ways...so answer is i or 2 - 10 years agoHelpfull: Yes(39) No(0)
- since we know that if a,b is prime factor of a number n.
and n=a^p*b^q
then number of divisor=(p+1)(q+1).
here given 10 divisor so possible factor of 10 is 1,10 or 2,5
so no of prime factors is either 1 or 2 - 10 years agoHelpfull: Yes(24) No(0)
- let us take 120 = 1x2x3x4x5x6x8x10x12x120
here it have 3 prime factors - 10 years agoHelpfull: Yes(2) No(26)
- c is the correct answer
- 10 years agoHelpfull: Yes(0) No(13)
- a,as a^p*b^q gives (p+1)*(q+1)
- 10 years agoHelpfull: Yes(0) No(3)
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