Hi,
Thanks for sharing an interesting question.
Let's say n = (p_1)^(a_1) x (p_2)^(a_2), since it has at
least 2 distinct prime factors.
Number of divisors of n = (a_1 + 1) x (a_2 + 1).
Can (a_1 + 1) x (a_2 + 1) = 13?
Thanks!
Cheers,
Wen Shih
Thank you Sir.
What I am thinking of is very similar, and draws from what I
have learnt in Finite Mathematics (Though with some twisting here
and there)
Suppose that an integer n is made up of j distinct primes, a_1,
a_2,....a_j, where j is an integer greater than 1.
Let n=a_1^m_1 X a_2^m_2 X .... a_j^m_j, where m_i's are all
positive integers.
Then the number of positive divisors of n will be (m_1 + 1) X
(m_2 + 1) X ... X (m_j + 1)
Notice that m_1+1, m_2+1, ..., m_j+1 are each at least 2,
whereas 13 can only be factorised into 1 X 13, and thus by
Fundamental theorem of Arithmetic, this is a
contradiction.
Hence, if an integer has at least two prime factors, it is not
possible to have 13 positive divisors, or any other prime amount of
positive divisors (E.g. 2, 3, 5, 7...)