A number has 101 composite factors. A number has 101 composite factors. How many prime factors at max A number could have ?
 A: Suppose $m = p_1^{a_1} ... p_n^{a_n}$ has evactly $101$ composite factors.
Then $101 + (1+n) = (a_1 + 1)(a_2+1) ... (a_n+1)$.
But the RHS is at least $2^n$ and it is easily checked that the inequality:
$102 + n \geq 2^n$
fails for $n \geq 7$. So there can be at most $6$ primes in the factorisation of $m$.
We now try to decompose the numbers $101 + (1+n)$ into a product of exactly $n$ integers for $n=1,2,3,4,5,6$, in order to see whether the $a_i$ can actually exist in each case.
We see that:
$108 = 2^2 \times 3^3$
$107$ is prime
$106 = 2\times 53$
meaning that the cases for $n=4,5,6$ cannot work.
However the number $105 = 3\times 5\times 7$ does have such a representation as a product of three numbers. Hence the biggest number of primes you may have in $m$ is $3$ in order to have exactly 101 composite factors.
Such a number is given by $m = p_1^2 p_2^4 p_3^6$ for any three different primes you wish.
As an aside, all such numbers $m$ must be of one of the following forms:
$p_1^2 p_2^4 p_3^6$
$p_1^7 p_2^{12}$
$p_1^3 p_2^{25}$
$p_1 p_2^{51}$
$p_1^{102}$
Where $p_1,p_2,p_3$ are distinct primes.
